M.C.Q

MCQ 11 Mark

In the adjoining figure, $AB \| CD$ and $AB \| EF$. The value of $x$ is:

A
$50^\circ$
B
$70^\circ $
C
$40^\circ$
✓
$60^\circ$

Answer

Correct option: D.

$60^\circ$

$\angle\text{FEC}+\angle\text{ECD}=180^\circ$ (Sum of $2$ supplementary angles is $180^\circ )$
$\angle\text{ECD}=\angle180^\circ-150^\circ=30^\circ$
$\angle\text{x}=\angle\text{BCE}=\angle\text{ECD}$
$\angle\text{x}=30^\circ+30^\circ=60^\circ.$

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MCQ 21 Mark

Side $BC$ of $\triangle\text{ABC}$ has been produced to Don left-hand side and to Eon right-hand side such that $\angle\text{ABD}=125^\circ$ and$\angle\text{ACE}=130^\circ$ then $\angle\text{A}=?$

A
$55^\circ $
B
$50^\circ$
✓
$75^\circ$
D
$65^\circ$

Answer

Correct option: C.

$75^\circ$

$\angle\text{ABD}+\angle\text{ABC}=180^\circ$(Linear Pair)
$\angle\text{ABC}=180^\circ-125^\circ=55^\circ$
$\angle\text{ACE}+\angle\text{ACB}=180^\circ$(Linear Pair)
$\angle\text{ACB}=180^\circ-130^\circ=50^\circ$
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$(Angle sum property)
$\angle\text{BAC}=180^\circ-50^\circ-55^\circ$
$\angle\text{BAC}=75^\circ$

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MCQ 31 Mark

In figure, $\text{PQ}||\text{RS},\angle\text{QPR}=70^\circ,\angle\text{ROT}=20^\circ$ find the value of $x.$

A
$20^\circ$
B
$70^\circ$
✓
$50^\circ$
D
$110^\circ$

Answer

Correct option: C.

$50^\circ$

$\text{PQ}||\text{RS}$
$\angle\text{QPR}=\angle\text{SRO}=70^\circ$ (Corresponding, Angle)
$\text{NOW IN}\triangle\text{RTO}$
$\text{x}+20^\circ=70^\circ$ (exterior angle)
$\text{x}=70^\circ-20^\circ$
$\text{x}=50^\circ$

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MCQ 41 Mark

Each angle of an equilateral triangle is:

A
$45^\circ$
B
$30^\circ$
✓
$60^\circ$
D
$90^\circ$

Answer

Correct option: C.

$60^\circ$

Let the angle of an equilateral triangle be $xox + x + x = 180^\circ $ (Angle sum property)
$3x = 180^\circ$
$x = 60^\circ .$

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MCQ 51 Mark

The angles of a triangle are in the ratio $2 : 3 : 4$. The largest angle of the triangle is:

A
$60^\circ$
B
$100^\circ $
C
$12^\circ$
✓
$80^\circ$

Answer

Correct option: D.

$80^\circ$

Suppose $\triangle\text{ABC}$ such that $\angle\text{A}:\angle\text{B}:\angle\text{C} = 2 : 3 : 4$
Let $\angle\text{A}=2\text{k},\angle\text{B}=3\text{k}$ and $\angle\text{C} = 4\text{k}$ where k is some constant
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Angle sum property)
$\Rightarrow 2k + 3k + 4k = 180^\circ$
$\Rightarrow 9k = 180^\circ$
$\Rightarrow k = 20^\circ $.

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MCQ 61 Mark

Which of the following pairs of angles are complementary?

✓
$25^\circ , 65^\circ$
B
$32.1^\circ , 47.9^\circ$
C
$70^\circ , 110^\circ$
D
$30^\circ , 70^\circ$

Answer

Correct option: A.

$25^\circ , 65^\circ$

Complementary angles always add up to $90^\circ$
$25^\circ + 65^\circ = 90^\circ$
Therefore this is the right option (by substitution).

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MCQ 71 Mark

If one of the angles of a triangle is $130^\circ $ then the angle between the bisectors of the other two angles can be:

A
$50^\circ$
B
$65^\circ$
C
$90^\circ$
✓
$155^\circ$

Answer

Correct option: D.

$155^\circ$

Let $\angle\text{A}=130^\circ$
In $\triangle\text{ABC},$ by angle sum property,
$\angle\text{B}+\angle\text{C}+\angle\text{A}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}+130^\circ=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}=50^\circ$
$\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}=25^\circ$
Now, in $\triangle\text{BOC},$
$\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\angle\text{BOC}=180^\circ$
$\Rightarrow25^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=155^\circ$

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MCQ 81 Mark

In Fig. if $l1 \| l2$, what is the value of $x?$

✓
$85^{\circ}$
B
$90^{\circ}$
C
$70^{\circ}$
D
$75^{\circ}$

Answer

Correct option: A.

$85^{\circ}$

Given that,
$l1 \| l2$
Let transversal $P$ and $Q$ cuts them
$\angle1=37^\circ$
$\angle4=58^\circ$
$\angle5=\text{x}^\circ$
$\angle1=\angle2=37^\circ$(Corresponding angles) (i)
$\angle2=\angle3$ (Vertically opposite angle)
$\angle3=37^\circ$
$\angle3+\angle4+\angle5=180^\circ$(Linear pair)
$37^{\circ}+58^{\circ}+x=180^{\circ}$
$x=85^{\circ}$.

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MCQ 91 Mark

In the given figure, the measure of $\angle1$ is:

A
$48^\circ$
✓
$42^\circ$
C
$138^\circ$
D
$158^\circ$

Answer

Correct option: B.

$42^\circ$

The two given angles are,"Vertically Opposite angles" which are known to be equal
Therefore, $\angle1=42^\circ.$

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MCQ 101 Mark

In the adjoining figure, what is the value of $y?$

A
$36$
✓
$54$
C
$63$
D
$72$

Answer

Correct option: B.

$54$

$AOB$ is a straight line.
$\therefore x^\circ + y^\circ 90^\circ = 180^\circ $
$\Rightarrow x + y = 90 .....(i)$
Since the angles around a point sum up to $360^\circ ,$
$\Rightarrow x^\circ + 90^\circ + y^\circ + 72^\circ + 3x^\circ = 360^\circ $
$\Rightarrow 4x + y = 198 .....(ii)$
Subtracting $(i)$ from $(ii)$, we get
$3x = 108 $
$\Rightarrow x = 36^\circ $
Substituting in $(i),$ we get
$y = 54^\circ $

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MCQ 111 Mark

In the adjoining figure, $\angle\text{a}$ and $\angle\text{g}$ are called:

A
Co-interior angles.
✓
Alternate exterior angles.
C
Corresponding angles.
D
Alternate interior angle.

Answer

Correct option: B.

Alternate exterior angles.

$\angle\text{a}$ and $\angle\text{g}$ are on alternate side and are exterior.

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MCQ 121 Mark

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $2 : 3$, then the greatest of two angles is:

A
$36^\circ $
✓
$108^\circ$
C
$72^\circ$
D
$54^\circ$

Answer

Correct option: B.

$108^\circ$

Let a and b are two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $2 : 3$ and we know that If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.Let common ratio is $x,$
$a = 2x$ and $b = 3x$
$a + b = 180^\circ$
$2x + 3x = 180^\circ$
$5x = 180^\circ$
$\text{x} = \frac{180^\circ}{5} = 36^\circ$
$x = 36^\circ$
$3x = 3 \times 36^\circ = 108^\circ .$

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MCQ 131 Mark

The number of lines that can pass through a given point is:

A
Only one.
✓
Infinity.
C
One.
D
Two.

Answer

Correct option: B.

Infinity.

As seen from the above image, any number of lines can be drawn through a given point.
Hence the answer may be given as $"$Infinity$"$.

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MCQ 141 Mark

In figure, which of the following statement must be true?
$(i) a + b = d + c$
$(ii) a + c + e = 180^\circ$
$(iii) b + f = c + e$

A
$(i)$ only
B
$(ii)$ only
C
$(iii)$ only
✓
$(ii)$ and $(iii)$ only

Answer

Correct option: D.

$(ii)$ and $(iii)$ only

From figure, we can see that
$\angle\text{a}^\circ+\angle\text{b}^\circ+\angle\text{c}^\circ=\angle\text{FOC}=180^\circ$
Also,
$\angle\text{b}^\circ=\angle\text{e}^\circ$ [Opposite angles]
So,
$\angle\text{a}^\circ+\angle\text{e}^\circ+\angle\text{c}^\circ=180^\circ$
$\Rightarrow (ii)$ is correct
Now,
$\angle\text{FOB}\neq\angle\text{DOB}$
$\Rightarrow\ \angle\text{a}^\circ+\angle\text{b}^\circ\neq\angle\text{d}^\circ+\angle\text{c}^\circ$
$\Rightarrow (i)$ is correct
Now,
$\angle\text{b}^\circ=\angle\text{e}^\circ$ and $\angle\text{f}^\circ=\angle\text{c}^\circ$ [Opposite angles are equal]
Thus,
$\angle\text{b}^\circ=\angle\text{f}^\circ=\angle\text{e}^\circ+\angle\text{c}^\circ$
$\Rightarrow (iii)$ is correct.

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MCQ 151 Mark

Write the correct answer in the following: In Fig. if $\text{AB}||\text{CD}||\text{EF},\text{PQ}||\text{RS},$ $\angle\text{RQD}=25^\circ$ and $\angle\text{CQP}=60^\circ,$ then $\angle\text{QRS}$ is equal to.

A
$85^\circ$
B
$135^\circ$
✓
$145^\circ$
D
$110^\circ$

Answer

Correct option: C.

$145^\circ$

Given, $\text{PQ}||\text{RS}$
$\angle\text{PQC}=\angle\text{BRC}=60^\circ$$\big[$alternate exterior angles and $\angle\text{PQC}=60^\circ$ (given)$\big]$ and $\angle\text{DQR}$
$=\angle\text{QRA}=25^\circ$ [alternate interior angles]
$$$\big[\angle\text{DQR}=25^\circ,\text{(given)}\big]$
$\angle\text{QRS}=\angle\text{QRA}+\angle\text{ARS}$
$=\angle\text{QRA}+\big(180^\circ-\angle\text{BRS}\big)$ [linear pair axiom]
$=25^\circ+180^\circ-60^\circ=205^\circ-60^\circ=145^\circ$

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MCQ 161 Mark

The measure of an angle is five times its comlement. The angle measure.

A
$25^\circ$
B
$35^\circ$
C
$65^\circ$
✓
$75^\circ$

Answer

Correct option: D.

$75^\circ$

Let the measure of the angle be $x^\circ ,$
So, its complement $= (90 - x)^\circ$
According to the given condition,
$x = 5(90 - x)$
$\Rightarrow x = 450 - 5x$
$\Rightarrow 6x = 450$
$\Rightarrow x = 75^\circ$
So, the angle measures $75^\circ .$

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MCQ 171 Mark

In the adjoining figure, the three lines $AB, CD$ and $EF$ all pass through the point O. If $\angle\text{EOB}=90^\circ$ and $x : y = 2 : 1$ then $\angle\text{BOD}$ and $\angle\text{COE}:$

✓
$30^\circ , 60^\circ$
B
$80^\circ , 20^\circ$
C
$45^\circ , 45^\circ$
D
$60^\circ , 60^\circ$

Answer

Correct option: A.

$30^\circ , 60^\circ$

$x + y + 90^\circ = 180^\circ $ (Linear Pair)
$2a + a + 90^\circ = 180^\circ $ (Since, $x : y = 2 : 1)$
$a = 30^\circ$
$\text{x}=\text{2a}=\angle\text{COE}=60^\circ$ (Vertically opposite angles)
$\text{y}=\angle\text{BOD}=30^\circ $ (Vertically opposite angles).

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MCQ 181 Mark

Measurement of reflex angle is:

✓
Between $180^\circ $ and $360^\circ $
B
$90^\circ $
C
Between $0^\circ $ and $90^\circ $
D
Between $90^\circ $ and $180^\circ$

Answer

Correct option: A.

Between $180^\circ $ and $360^\circ $

Let $x$ be the angle then its reflex angle is $360^\circ - x$ and in any triangle, the angle lies between $0$ to $180^\circ .$

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MCQ 191 Mark

In the given figure $x = 30^\circ $, the value of $Y$ is:

A
$36^\circ $
B
$10^\circ$
✓
$40^\circ$
D
$45^\circ$

Answer

Correct option: C.

$40^\circ$

In the given figure we have
$3Y + 2X = 180^\circ $ (Linear - Pair)
$X = 30^\circ$
$3Y + 2 \times 30^\circ = 180^\circ$
$3Y + 60^\circ = 180^\circ$
$3|Y = 180^\circ - 60^\circ$
$3Y = 120^\circ$
$\text{Y}=\frac{120^\circ}{3}$
$Y = 40^\circ .$

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MCQ 201 Mark

In the given figure, straight lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{BOD}=130^\circ$ then $\angle\text{AOD}=?$

A
$65^\circ $
✓
$115^\circ$
C
$110^\circ$
D
$125^\circ$

Answer

Correct option: B.

$115^\circ$

$\angle\text{AOC}+\angle\text{BOD}=1306^\circ$ (given)But $\angle\text{AOC}=\angle\text{BOD}$ (Vartically Opposite angles)
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Since $COD$ is a straight line,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOD}=115^\circ$

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MCQ 211 Mark

The measure of an angle is five times its complement. The angle measures.

A
$25^\circ$
✓
$75^\circ$
C
$65^\circ$
D
$35^\circ$

Answer

Correct option: B.

$75^\circ$

Let the measure of the required angle be $x^\circ$
Then, the measure of its complement will be $(90 − x)^\circ$
Therefore, $x = 5 (90 - x)$
$\Rightarrow x = 450 - 5x$
$\Rightarrow 6x = 450$
$\Rightarrow x = 75^\circ $.

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MCQ 221 Mark

$A, B, C$ are the three angles of a triangle. If $A - B = 15^\circ $ and $B - C = 30^\circ $, then angles$ A, B, C$ are respectively:

✓
$80^\circ , 65^\circ , 35^\circ$
B
$65^\circ , 80^\circ , 35^\circ$
C
$80^\circ , 35^\circ , 65^\circ$
D
$35^\circ , 65^\circ , 80^\circ$

Answer

Correct option: A.

$80^\circ , 65^\circ , 35^\circ$

Since $ABC$ is a triangle
$A + B + C = 180^\circ$ (Angle sum property) $(i)$
$A - B = 15^\circ$
$A = 15^\circ + B (ii)$
$B - C = 30^\circ$
$C = B - 30^\circ (iii)$
From $(i)$ equation
$15^\circ + B + B + B - 30^\circ = 180^\circ$
$B = 65^\circ$
From equation $(ii)$ and $(iii)$
$A = 15^\circ + B = 15^\circ + 65^\circ = 80^\circ$
$C = B - 30^\circ = 65^\circ - 30^\circ = 35^\circ .$

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MCQ 231 Mark

In the given figure, $AB \| CD$, If $\angle\text{APQ}=70^\circ$ and $\angle\text{PRD}=120^\circ,$ then $\angle\text{QPR}=?$

A
$35^\circ$
B
$40^\circ$
C
$60^\circ$
✓
$50^\circ$

Answer

Correct option: D.

$50^\circ$

$\angle\text{APQ}=\angle\text{PQR}=70^\circ$ (Alternate interior angles)
$\angle\text{PRQ}+\angle\text{PRD}=180^\circ$ (Linear Pair)
$\angle\text{PRQ}=180^\circ-120^\circ=60^\circ$
In $\triangle\text{PQR}$
$\angle\text{PQR}+\angle\text{PRQ}+\angle\text{QPR}=180^\circ$ (Angle sum property)
$\angle\text{QPR}=180^\circ-70^\circ-60^\circ=50^\circ.$

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MCQ 241 Mark

The angles of a triangle in ascending order are $x, y, z$ and $y - x = z - y = 10^{\circ}$. The smallest angles is:

A
$60^{\circ}$
✓
$50^{\circ}$
C
$70^{\circ}$
D
$40^{\circ}$

Answer

Correct option: B.

$50^{\circ}$

$x+y+z=180^{\circ} \text { (i) (Angle sum property) }$
$y-x=10^{\circ}$
$y-10^{\circ}=x \text { (ii) }$
$z-y=10^{\circ}$
$z=10^{\circ}+y \text { (iii) }$
On putting the value of $x$ and $z$ in equation (i)
$y-10^{\circ}+y+10^{\circ}+y=180^{\circ}$
$y=60^{\circ}$
$x=50^{\circ}(\text { From equation ii })$
$z=70^{\circ}(\text { From equation iii) }$
Smallest angle is $x=50^{\circ}$.

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MCQ 251 Mark

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $5 : 4$, then the smaller of the two angles is:

A
$120^\circ$
B
$60^\circ$
✓
$80^\circ$
D
$100^\circ$

Answer

Correct option: C.

$80^\circ$

We know that sum of two interior angles on the same side of a transversal intersecting two parallel lines is $180^\circ .$
Let the common ratio is $x.$
So the angles are$ 5x, 4x.$
So, $5x + 4x = 180^\circ$
$9x = 180^\circ$
$\text{x}=\frac{180^\circ}{9}$
$x = 20^\circ$
So the angles are $5x = 100^\circ .$
$4x = 80^\circ$
So smallest angle is $80^\circ .$

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MCQ 261 Mark

In Fig. $AB \| CD \| EF$ and $GH \| KL$. The measure of $\angle\text{HKL}$ is:

A
$85^\circ$
B
$135^\circ$
C
$215^\circ$
✓
$145^\circ$

Answer

Correct option: D.

$145^\circ$

Given, $AB \| CD \| EF$ and $GH \|$ KLProduce $HG$ to $M$ and KL to $N$
$\angle\text{MHD}$ and $\angle\text{CHG}=60^\circ$ (Vertically opposite angle)
Since, $MG \| NL$ and transversal cuts them
So, $\angle\text{MHD}+\angle1=180^\circ$ (Interior angles)
$60^\circ+\angle1=180^\circ$
$\angle1=120^\circ$
$\angle3=\angle\text{HKD}=25^\circ$ (Alternate angles) $(i)$
$\angle1=\angle\text{MKL}=120^\circ$ (Corresponding angles) $(ii)$
Now, $\angle\text{HKL}=\angle3+\angle\text{MKL}$
$= 25^\circ + 120^\circ$
$= 145^\circ .$

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MCQ 271 Mark

In the adjoining figure, if $QP \| RT$, then $x$ is equal to:

A
$55^\circ$
✓
$75^\circ$
C
$65^\circ$
D
$70^\circ$

Answer

Correct option: B.

$75^\circ$

$\angle\text{QPR}=\angle\text{PRT}=40^\circ$ (Alternate interior angles)
In $\triangle\text{QPR}$
$\angle\text{PQR}+\angle\text{QPR}+\angle\text{PRQ}=180\circ $ (Angle sum property)
$65^\circ+40^\circ+\text{x}^\circ=180^\circ$
$\text{x}^{\circ}=180^\circ-40^\circ-65^\circ$
$\text{x}^{\circ}=75$

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MCQ 281 Mark

In the given $x =?$

A
$\alpha-\beta-\gamma$
✓
$\alpha+\beta+\gamma$
C
$\alpha+\beta-\gamma$
D
$\alpha+\gamma-\beta$

Answer

Correct option: B.

$\alpha+\beta+\gamma$

$OBCA$ is a quadrilateral$\angle\text{OAC}+\angle\text{BOA}+\angle\text{ACB}+\angle\text{CBO}=360^\circ$
$\gamma+\beta+\angle\text{ACB}+\alpha=360^\circ$
$\angle\text{ACB}=360^\circ-\gamma-\beta-\alpha$
$\text{x}=360^\circ-\angle\text{ACB}$
$\text{x}=\alpha+\beta+\gamma.$

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MCQ 291 Mark

Write the correct answer in the following: The angles of a triangle are in the ratio $5 : 3 : 7$ The triangle is.

✓
An acute angled triangle.
B
An obtuse angled triangle.
C
A right triangle.
D
An isosceles triangle.

Answer

Correct option: A.

An acute angled triangle.

Let the angles of the triangle be $5x, 3x$ and $7x$.As the sum of the angles of a triangle is $180^\circ $ then
$5x + 3x + 7x = 180^\circ$
$\Rightarrow 15x = 180^\circ$
$\Rightarrow x = 180^\circ ÷ 15 = 12^\circ$
Therefore, the angle of the triangle are:
$5 \times 12^\circ , 3 \times 12^\circ $ and $7 \times 12^\circ , i.e., 60^\circ , 36^\circ $ and $84^\circ$
As the measure of each angle of the triangle is less than $90^\circ $, so the angles of triangle are acute angles.
Therefore, the triangle is an acute angled triangle.
Hence, $(a)$ is the correct answer.

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MCQ 301 Mark

In the given figure, $\angle\text{OEB} = 75^\circ,\angle\text{OBE}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ Then $\angle\text{ODC}=?$

✓
$30^\circ$
B
$25^\circ$
C
$35^\circ$
D
$20^\circ$

Answer

Correct option: A.

$30^\circ$

In $\triangle\text{OEB}$
$\angle\text{OEB}+\angle\text{EBO}+\angle\text{BOE}=180^\circ$ (Angle sum property)
$75^\circ+55\circ+\angle\text{BOE}=180^\circ$
$\angle\text{BOE}=50^\circ$
$\angle\text{BOE}=\angle\text{COD}=50^\circ$ (vertically opposite angle)
In $\triangle\text{ODC}$
$\angle\text{ODC}+\angle\text{DOC}+\angle\text{DCO}=180^\circ$
$\angle\text{ODC}=180^\circ-100^\circ-50^\circ$
$\angle\text{ODC}=30^\circ$

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MCQ 311 Mark

An angle is one fifth of its supplement. The measure of the angle is:

A
$15^\circ$
✓
$30^\circ$
C
$75^\circ$
D
$150^\circ$

Answer

Correct option: B.

$30^\circ$

Let the measure of the angle be $x^\circ .$
So, its supplement $= (180^\circ - x)$
According to the given condition,
$\text{x}=\frac{1}{5}(180^\circ-\text{x)}$
$\Rightarrow5\text{x}=180-\text{x}$
$\Rightarrow6\text{x}=180$
$\Rightarrow\text{x}=30^\circ$

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MCQ 321 Mark

In Fig., the value of $x$ is:

A
$15^\circ$
B
$12^\circ$
C
$8^\circ$
✓
$20^\circ$

Answer

Correct option: D.

$20^\circ$

Let,$AB, CD$ and $EF$ intersect at $O$
$\angle\text{COB}=\angle\text{AOD}$ (Vertically opposite angle)
$\angle\text{AOD}=3\text{x}+10\text{ (i)}$
$\angle\text{AOE}+\angle\text{AOD}+\angle\text{DOF}=180^\circ$ (Linear pair)
$x + 3x + 10^\circ + 90^\circ = 180^\circ$
$4x + 100^\circ = 180^\circ$
$4x = 80^\circ$
$x = 20^\circ .$

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MCQ 331 Mark

Two straight lines $AB$ and $CD$ cut each other at $O$. If $\angle\text{BOD}=63^\circ,$ then $\angle\text{BOC}=$

A
$63^\circ$
✓
$117^\circ$
C
$17^\circ$
D
$153^\circ$

Answer

Correct option: B.

$117^\circ$

$\angle\text{BOD}$ and $\angle\text{BOC}$ from a linear pair.
$\therefore\ \angle\text{BOD}+\angle\text{BOC}=180^\circ$
$\Rightarrow\ 63^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\ \angle\text{BOC}=117^\circ$

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MCQ 341 Mark

In a given figure, if $AB \| CD \| EF, PQ \| RS$, $\angle\text{RQD}=25^\circ$ and $\angle\text{CQP}= 60^\circ,$ then $\angle\text{QRS}$ is equal to:

✓
$145^\circ $
B
$110^\circ$
C
$85^\circ$
D
$135^\circ$

Answer

Correct option: A.

$145^\circ $

Given, $PQ \| RS$
$\angle\text{PQC}=\angle\text{BRS}=60^\circ$ [alternate exterior angles and $\text{PQC}=60^\circ$ (given)] and $\angle\text{DQR}=\angle\text{QRA}=25^\circ$ [alternate interior angles]
$[\angle\text{DQR}=25^\circ, \text{ given}]$
$\angle\text{QRS}=\angle\text{QRA}+\angle\text{ARS}$
$=\angle\text{QRA}+(180^\circ–\angle\text{BRS})$ [linear pair axiom]
$=25^\circ+180^\circ-60^\circ=205^\circ- 60^\circ=145^\circ.$

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MCQ 351 Mark

The angles of a triangle are in the ratio $2 : 3 : 4$. The largest angle of the triangle is:

A
$120^\circ$
B
$100^\circ$
✓
$80^\circ$
D
$60^\circ$

Answer

Correct option: C.

$80^\circ$

By angle sum property,
$2x + 3x + 4x = 180^\circ$
$\Rightarrow 9x = 180^\circ$
$\Rightarrow x = 20^\circ$
Hence, largest angle $= 4x = 4(20^\circ ) = 80^\circ$

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MCQ 361 Mark

In the given figure, the measure of $\angle\text{a}$ is:

A
$150^\circ$
✓
$30^\circ$
C
$15^\circ$
D
$50^\circ$

Answer

Correct option: B.

$30^\circ$

In the given figure
$150^\circ\angle\text{a}=180^\circ$ (linear - pair)
$\angle\text{a}=180^\circ=150^\circ$
Therefore,
$\angle\text{a}=30^\circ$

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MCQ 371 Mark

The angle which is equal to 8 times its complement is:

A
$88^\circ$
B
$72^\circ$
✓
$80^\circ$
D
$90^\circ$

Answer

Correct option: C.

$80^\circ$

We know that two angles, whose sum is $90^\circ $, are called the complementary angle.
Let one angle be x then its complementary angle be 8x,
$x + 8x = 90^\circ$
$9x = 90^\circ$
$x = 10^\circ$
Its complementary angle is $8 \times 10 = 80^\circ .$

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MCQ 381 Mark

Given $\angle\text{POR}=3\text{x}$ and $\angle\text{QOR}=2\text{x}+10^\circ.$ If $\angle\text{POQ}$ is a straight line, then the value of $x$ is:

A
$30^\circ $
B
$36^\circ $
✓
$34^\circ $
D
None of these

Answer

Correct option: C.

$34^\circ $

Given,
$POQ$ is a straight line
$\angle\text{POR}+\angle\text{QOR}=180^\circ$ (Linear pair)
$3x + 2x + 10^\circ = 180^\circ$
$5x = 170^\circ$
$x = 34^\circ .$

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MCQ 391 Mark

In the adjoining figure, $AB \| CD$ and $AB \| EF$. If $EA$ $\bot$ $BA$ and $\angle\text{BEF}$ then the values of $x, y$ and $z:-$

✓
$125^\circ , 125^\circ , 35^\circ$
B
$60^\circ , 60^\circ , 60^\circ$
C
$120^\circ , 130^\circ , 25^\circ$
D
$35^\circ , 125^\circ , 120^\circ$

Answer

Correct option: A.

$125^\circ , 125^\circ , 35^\circ$

$x + 55 = 180^\circ$ (Sum of supplementary angles or co-interior angles)$x = 125^\circ $
$x = y = 125^\circ $ (Corresponding angles)
$\text{z}+\angle\text{EAB}$ (Exterior angle property)
$\text{z}=125^\circ-90^\circ=35^\circ$

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MCQ 401 Mark

In the given figure, $AB \| CD$ and $O$ is a point joined with $B$ and $D$, as shown in the figure such that $\text{ABO}=35^\circ$ and $\angle\text{CDO}=40^\circ$ Reflex $\angle\text{BOD}=?$

A
$265^\circ$
✓
$285^\circ$
C
$275^\circ$
D
$255^\circ$

Answer

Correct option: B.

$285^\circ$

$\angle\text{ABO}+\angle\text{BOE}=180^\circ$ (Sum of supplementary angles)
$\angle\text{BOE}=180^\circ-35^\circ=145^\circ$
$\angle\text{CDO}+\angle\text{DOE}=180^\circ$ (Sum of supplementary angles)
Reflex of $\angle\text{BOD}=\angle\text{BOE}+\angle\text{DOE}=145^\circ+140^\circ$
Reflex of $\angle\text{BOD}=285^\circ.$

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MCQ 411 Mark

In the given figure, $AOB$ is a straight line. The value of $x$ is:

✓
$15$
B
$20$
C
$25$
D
$12$

Answer

Correct option: A.

$15$

It is given that, $A O B$ is a straight line.
$\therefore$ $60^\circ + (5x^\circ + 3x^\circ ) = 180^\circ $ (Linear pair)
$\Rightarrow 8x^\circ = 180^\circ - 60^\circ = 120^\circ$
$\Rightarrow x^\circ = 15^\circ$
Thus, the value of $x$ is $15.$

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MCQ 421 Mark

For what value of $x$ shall we have $l \| m?$

A
$x = 60^\circ$
✓
$x = 50^\circ$
C
$x = 70^\circ$
D
$x = 45^\circ$

Answer

Correct option: B.

$x = 50^\circ$

$(2x - 30)^\circ = (x + 20)^\circ$ (corresponding angle)
$2x -30^\circ = x + 20^\circ$
$2x - x = 30^\circ + 20^\circ$
$x = 50^\circ .$

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MCQ 431 Mark

In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=4\text{x}^\circ$ and $\angle\text{BOC}=5\text{x}^\circ$ then $\angle\text{AOC}=?$

A
$100^\circ$
B
$40^\circ$
✓
$80^\circ$
D
$60^\circ$

Answer

Correct option: C.

$80^\circ$

We have,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$ [Since $AOB$ is a straight line]
$\Rightarrow 4x + 5x = 180^\circ$
$\Rightarrow 9x = 180^\circ$
$\Rightarrow x = 20^\circ$
$\therefore\angle\text{AOC}=4\times20^\circ=80^\circ.$

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MCQ 441 Mark

In the given figure, $AB \| DC$, $\angle\text{BAD}=90^\circ,\angle\text{CBD}=28^\circ$ and $\angle\text{BCE}=65^\circ.$ Then $\angle\text{ABD}=?$

A
$32^\circ$
✓
$37^\circ$
C
$43^\circ$
D
$53^\circ$

Answer

Correct option: B.

$37^\circ$

In $\triangle\text{DBC}$
$\angle\text{BCE}=\angle\text{DBC}+\angle\text{BDC}$ (Exterior angle property)
$65^\circ=28^\circ+\angle\text{BDC}$
$\text{BDC}=37^\circ$
As, $AB$ is parallel to $CD$
$\angle\text{ABD}=\angle\text{BDC}=37^\circ$ (Alternate interior angle).

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MCQ 451 Mark

In the given below figure, the measure of $\angle\text{AED}$ is:

A
$140^\circ$
✓
$130^\circ$
C
$120^\circ$
D
$110^\circ$

Answer

Correct option: B.

$130^\circ$

in $\triangle\text{ABC}$$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$ (Angle sum property)
$\angle\text{ACB}=180^\circ-25^\circ-45^\circ$
$\angle\text{ACB}=110^\circ$
$\angle\text{ACB}+\angle\text{ACD}=180^\circ$ (Linear pair)
$\angle\text{ACD}=180^\circ-110^\circ=70^\circ$
In $\triangle\text{CED}$
$\angle\text{AED}+\angle\text{EDC}+\angle\text{EDC}$ (Exterior angle is equal to sum of its two interior opposite angles)
$\angle\text{AED}=60^\circ+75^\circ=130^\circ.$

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MCQ 461 Mark

In the adjoining figure, $y =?$

A
$36^\circ$
B
$63^\circ$
C
$72^\circ$
✓
$54^\circ$

Answer

Correct option: D.

$54^\circ$

We have,$3x + 72 = 180^\circ $[$\because$ AOB is a straight line]
$\Rightarrow 3x = 108$
$\Rightarrow x = 36$
Also,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$ [$\because$ $AOB$ is a straight line]
$\Rightarrow 36^\circ + 90^\circ + y = 180^\circ$
$\Rightarrow y = 54^\circ .$

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MCQ 471 Mark

In figure, if line segment AB is parallel to the line segment $CD$, what is the value of $y?$

A
$12$
B
$15$
C
$18$
✓
$20$

Answer

Correct option: D.

$20$

From figure,
$\angle\text{ABD}+\angle\text{EBD}=180^\circ$
$\Rightarrow\ \angle\text{EBD}=180^\circ-\angle\text{ABD}\dots(1)$
Now,
$\angle\text{ABD}=\text{y}^\circ+2\text{y}^\circ+\text{y}^\circ$
$\Rightarrow\ \angle\text{ABD}=4\text{y}^\circ\dots(2)$
Substituting (2) in (1), we have
$\angle\text{EBD}=180^\circ-4\text{y}^\circ$
Now,
$\angle\text{EBD}=\angle\text{BDC}$ [Alternate angles]
$\Rightarrow\ 180^\circ-4\text{y}^\circ=5\text{y}^\circ$
$\Rightarrow\ 180^\circ=9\text{y}^\circ$
$\Rightarrow\ \text{y}^\circ=20^\circ$

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MCQ 481 Mark

In Fig., which of the following statements must be true?
$i. a + b = d + c$
$ii. a + c + e = 180^\circ$
$iii. b + f = c + e$

A
$(iii)$ only
B
$(i)$ only
C
$(ii)$ only
✓
$(ii)$ and $(iii)$ only

Answer

Correct option: D.

$(ii)$ and $(iii)$ only

Let $AB, CD$ and $EF$ intersect at $O$
$\angle\text{AOD}=\angle\text{COB} ($Vertically opposite angle$)$
$b = e (i)$
$\angle\text{EOC}=\angle\text{DOF} ($Vertically opposite angle$)$
$f = c (ii)$
Adding $(i)$ and $(ii),$ we get
$b + f = c + e (iii)$
Now,
$\angle\text{AOD}+\angle\text{EOC}+\angle\text{COB}=180^\circ$
$a + f + e = 180^\circ$
$a + c + e = 180^\circ $ [From $(ii)].$

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MCQ 491 Mark

The number of angles formed by a transversal with a pair of parallel lines are:

✓
$8$
B
$6$
C
$3$
D
$4$

Answer

Correct option: A.

$8$

As we can see there are $4$ angles formed at every point of intersection thus giving a total of $8$ angles.

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MCQ 501 Mark

If $\angle\text{A}=4\angle\text{B} = 6\angle\text{C},$ then $A : B : C?$

A
$6 : 4 : 3$
✓
$12 : 3 : 2$
C
$2 : 3 : 4$
D
$3 : 4 : 6$

Answer

Correct option: B.

$12 : 3 : 2$

Let A be x$\text{B}=\frac{1}{4}\text{x}$
$\text{C}=\frac{1}{6}\text{x}$
$A : B : C$
$\text{x}=\frac{1}{4}\text{x}:\frac{1}{6}\text{x}$
$LCM$ of $4$ and $6$ is $12$
$12 : 3 : 2.$

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MCQ 511 Mark

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:

A
An isosceles triangle.
B
An obtuse triangle.
C
An equilateral triangle.
✓
A right triangle.

Answer

Correct option: D.

A right triangle.

In a right triangle, one angle is 90° and the sum of acute angles of a right triangle is $90^\circ .$

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MCQ 521 Mark

Given $\angle\text{POR}=3\text{x}$ and $\angle\text{QOR}=2\text{x}+10^\circ.$If $POQ$ is a straight line, then the value of $x$ is:

A
$30^\circ$
✓
$34^\circ$
C
$36^\circ$
D
None of these

Answer

Correct option: B.

$34^\circ$

$\angle\text{POR}=3\text{x}$ and $\angle\text{QOR}=2\text{x}+10^\circ$
From figure, we can see that $\angle\text{POR}$ and $\angle\text{QOR}$ are two adjacent angles and are supplement.
$\Rightarrow\ \angle\text{POR}+\angle\text{QOR}=180^\circ$
$\Rightarrow\ 3\text{x}+2\text{x}+10^\circ=180^\circ$
$\Rightarrow\ 5\text{x}=170^\circ$
$\Rightarrow\ \text{x}=34^\circ$

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MCQ 531 Mark

In the adjoining figure $\angle\text{QPR}=62^\circ$ and $\angle\text{PRQ}=64^\circ$ If $OQ$ and $OR$ and bisectors of $\angle\text{PQR}$ and $\angle\text{PRQ}$ respectively, then $\angle\text{OQR}$ and $\angle\text{QOR}:$

A
$121^\circ , 20^\circ$
✓
$27^\circ , 121^\circ$
C
$20^\circ , 80^\circ$
D
$26^\circ , 124^\circ$

Answer

Correct option: B.

$27^\circ , 121^\circ$

In $\triangle\text{PQR}$
$\angle\text{QPR}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$ (Angle sum property)
$\angle\text{PQR}=180^\circ-62^\circ-64^\circ$
$\angle\text{PQR}=54^\circ$
$\angle\text{ORQ}=32^\circ$ (OR is a bisector)
$\angle\text{OQR}=27^\circ$ (OR is a bisector)
In $\triangle\text{OQR}$
$\angle\text{OQR}+\angle\text{ORQ}+\angle\text{QOR}=180^\circ$ (Angle sum property)
$\angle\text{QOR}=180^\circ-32^\circ-27^\circ=121^\circ$

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MCQ 541 Mark

Two lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ,$ then $\angle\text{AOC}=$

A
$180^\circ$
B
$70^\circ$
C
$80^\circ$
✓
$90^\circ$

Answer

Correct option: D.

$90^\circ$

Given that,
$AB$ and $CD$ intersect at $O$
$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ\text{(i)}$
$\angle\text{COB}+\angle\text{BOD}=180^\circ$ (Linear pair) $(ii)$
Using $(ii)$ in $(i)$, we get
$\angle\text{AOC}+180^\circ=270^\circ$
$\text{AOC}=90^\circ.$

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MCQ 551 Mark

Given two distinct points $P$ and $Q$ in the interior of $\angle\text{ABC},$ then $\overrightarrow{\text{AB}}$ will be:

A
In the exterior of $\angle\text{ABC}$
✓
On the $\angle\text{ABC}$
C
On the both sides of $\overrightarrow{\text{BA}}$
D
In the interior of $\angle\text{ABC}$

Answer

Correct option: B.

On the $\angle\text{ABC}$

$\overrightarrow{\text{AB}}$ is a line which from $\angle\text{ABC}$ and it is the part of $\triangle\text{ABC}.$

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MCQ 561 Mark

In the given figure, the measure of $\angle\text{ABC}$ is:

A
$20^\circ$
✓
$80^\circ$
C
$60^\circ$
D
$100^\circ$

Answer

Correct option: B.

$80^\circ$

$\angle\text{A}=20^\circ$ (Vertical opp Angle)
In $\angle\text{ABC}$
The exterior angle so formed is equal to the sum of the two interior opposite angles.
$\angle\text{A}+\angle\text{ABC}=100^\circ$
$20^\circ+\angle\text{ABC}=100$
$\angle\text{ABC}=100^\circ-20^\circ$
$\angle\text{ABC}=180^\circ.$

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MCQ 571 Mark

The exterior angle of a triangle is equal to the sum of two:

✓
Interior opposite angles.
B
Alternate angles.
C
Exterior angles.
D
Interior angles.

Answer

Correct option: A.

Interior opposite angles.

$\angle1+\angle2+\angle3=180^\circ$ (Angle sum property) $(a)$
$\angle3+\angle4=180^\circ$ (Linear pair) $(b)$
On equating equations a and b, we get
$\angle1+\angle2=\angle4.$

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MCQ 581 Mark

In the given figure, $\angle\text{BAC}=30^\circ,\angle\text{ABC}=50^\circ$ and $\angle\text{CDE}=40^\circ.$ Then $\angle\text{AED}=?$

✓
$120^\circ$
B
$110^\circ$
C
$80^\circ$
D
$100^\circ$

Answer

Correct option: A.

$120^\circ$

In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{BCA}=180^\circ$ (Angle sum property)
$50^\circ+30^\circ+\angle\text{BCA}=180^\circ$
$\angle\text{BCA}=100^\circ$
In $\triangle\text{ECD}$
$\angle\text{ECD}+\angle\text{CDE}+\angle\text{CED}=180^\circ$ (Angle sum property)
$180^\circ-\angle\text{BCA}+40^\circ+\angle\text{CED}=180^\circ$
$\angle\text{CED}=100^\circ-40^\circ=60^\circ$
$\angle\text{CED}+\angle\text{AED}=180^\circ$ (Linear Pair)
$\angle\text{AED}=180^\circ-60^\circ=120^\circ.$

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MCQ 591 Mark

In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=(3\text{x}-10)^\circ,\angle\text{COD}=50^\circ$ and $\angle\text{BOD}=(\text{x}+20)^\circ$ then $\angle\text{AOC}=?$

A
$50^\circ$
B
$60^\circ$
✓
$80^\circ$
D
$40^\circ$

Answer

Correct option: C.

$80^\circ$

We have,$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$ [Since $AOB$ is a straight line]
$\Rightarrow 3x - 10 + 50 + x + 20 = 180$
$\Rightarrow 4x = 120$
$\Rightarrow x = 30$
$\therefore\angle\text{AOC}=[3\times30−10]^\circ$
$\Rightarrow\angle\text{AOC}=80^\circ.$

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MCQ 601 Mark

In Fig., If $AB \| CD$, then the value of $x$ is:

✓
$30$
B
$12$
C
$15$
D
$20$

Answer

Correct option: A.

$30$

In the given figure we have,
$120^\circ + x + x = 180^\circ$
$\Rightarrow 2x = 180^\circ - 120^\circ$
Or, $\text{x}=\frac{60}{2}=30^\circ.$

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MCQ 611 Mark

In the given figure, $AB \| CD$. If $\text{BAO}=60^\circ$ and $\angle\text{OCD}=110^\circ,$ then $\angle\text{AOC}=?$

A
$40^\circ$
✓
$50^\circ$
C
$60^\circ$
D
$70^\circ$

Answer

Correct option: B.

$50^\circ$

Construction: Extend the line $CD$ such that it intersect $AO$ and is parallel to $AB$
$x = 60^\circ $ (Corresponding angles)
$x + y + 180^\circ - 110^\circ = 180^\circ $ (Angle sum property)
$y = 110^\circ - 60^\circ = 50^\circ .$

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MCQ 621 Mark

In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=(3\text{x} +10)^\circ$ and $\angle\text{BOC}=(4\text{x}-26)^\circ,$ then $\angle\text{BOC}=?$

A
$96^\circ$
✓
$86^\circ$
C
$76^\circ$
D
$106^\circ$

Answer

Correct option: B.

$86^\circ$

Since $AOB$ is a straight line,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow(3\text{x}+10)+(4\text{x}-26)=180^\circ$
$\Rightarrow7\text{x}-16=180^\circ$
$\Rightarrow7\text{x}=196$
$\Rightarrow\text{x}=28$
So, $\angle\text{BOC}=4 \text{x}-26=4(28)-26=86^\circ$

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MCQ 631 Mark

In figure, $AB$ and $CD$ are parallel to each other. The value of $x$ is:

A
$90^\circ$
B
$140^\circ$
✓
$100^\circ$
D
$120^\circ$

Answer

Correct option: C.

$100^\circ$

let us draw a line from point E parallel to line $AB, CD$
$\text{X}=\angle1+\angle2$
$AB \| EF$
$\angle1+120^\circ=180^\circ$ (Co - interior angle)
$\angle1=180^\circ-120^\circ$
$\angle1=60^\circ$
$CD \| EF$
$\angle2+140^\circ=180^\circ$ (Co - interior angle)
$\angle2=180^\circ-140^\circ$
$\angle1=40^\circ$
$\text{X}=\angle1+\angle2$
$\text{X}=60^\circ+40^\circ.$

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MCQ 641 Mark

The number of angles formed by a transversal with a pair of parallel lines are.

✓
$8$
B
$4$
C
$6$
D
$3$

Answer

Correct option: A.

$8$

As we can see there are $4$ angles formed at every point of intersection thus giving a total of $8$ angles.

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MCQ 651 Mark

The sides $BC, BA$ and $CA$ of $\triangle\text{ABC}$ have been produced to $D, E$ and $F$ respectively, as shown in the give figure, Then, $\angle\text{B}?$

A
$35^\circ$
B
$65^\circ$
C
$55^\circ$
✓
$75^\circ$

Answer

Correct option: D.

$75^\circ$

$\angle\text{FAE}=\text{BAC}(\text{VOA})$$\angle\text{BAC}=35^\circ$
$\angle\text{ACB}+\angle\text{ACD}=180^\circ$ (Linear Pair)
$\angle\text{ACB}+110^\circ=180^\circ$
$\angle\text{ACB}=180^\circ-110^\circ$
$\angle\text{ACB}=70^\circ$
$\angle\text{BAC}+\angle\text{B}+\angle\text{ACB}=180^\circ$
$35^\circ+\angle\text{B}+70^\circ=180^\circ$
$\angle\text{B}+105^\circ=180^\circ$
$\angle\text{B}=180^\circ-105^\circ$
$\angle\text{B}=75^\circ.$

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MCQ 661 Mark

If two angles are supplementary and the larger is $20^\circ $ less then three times the smaller, then the angles are:

A
$72\frac{1}{2}^\circ,17\frac{1}{2}^\circ$
B
$140^\circ,40^\circ$
C
$62\frac{1}{2}^\circ,27\frac{1}{2}^\circ$
✓
$130^\circ,50^\circ$

Answer

Correct option: D.

$130^\circ,50^\circ$

Let the two supplementary angles be $x^\circ $ and $180^\circ - x^\circ $
Let $180^\circ - x = 3x 20^\circ$
$4x = 200^\circ$
$x = 50^\circ$
So the angles are $50^\circ $ and $130^\circ .$

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MCQ 671 Mark

In figure, $PQ \| RS$, $\angle\text{AEF}=95^\circ,\angle\text{BHS}=110^\circ$ and $\angle\text{ABC}=\text{x}^\circ.$ Then the value of $x$ is:

A
$15^\circ$
✓
$25^\circ$
C
$70^\circ$
D
$35^\circ$

Answer

Correct option: B.

$25^\circ$

From figure,
$\angle\text{AEF}=\angle\text{EGH}$ [Corresponding angles]
$\Rightarrow\ \angle\text{EGH}=\angle\text{AEF}=95^\circ$
Also,
$\angle\text{BGH}+\angle\text{EGH}=180^\circ$
$\Rightarrow\ \angle\text{BGH}=180^\circ-\angle\text{EGH}=180^\circ-95^\circ$
$=85^\circ$
$\angle\text{BHS}=110^\circ$
Now,
$\angle\text{BHG}+\angle\text{BHS}=180^\circ$
$\Rightarrow\ \angle\text{BHG}=180^\circ-\angle\text{BHS}=180^\circ-110^\circ$
$=70^\circ$
Now, in $\triangle\text{BHG}$
$\angle\text{BGH}+\angle\text{BGH}+\text{x}=180^\circ$ [Sum of all angles of a $\triangle$ is 180°]
$\Rightarrow\ 85^\circ+70^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow\ \text{x}^\circ=180^\circ-155^\circ=25^\circ$

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MCQ 681 Mark

In the given figure, $\text{AM }\bot\text{ BC}$ and AN is the bisector of $\angle\text{A}.$ If $\angle\text{ABC}=70^\circ$ and $\angle\text{ACB}=20^\circ,$ then $MAN =?$

A
$20^\circ$
B
$30^\circ$
C
$15^\circ$
✓
$25^\circ$

Answer

Correct option: D.

$25^\circ$

In $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ (Angle sum property)
$\angle\text{BAC}=180^\circ-70^\circ-20^\circ$
$\angle\text{BAC}=90^\circ$
In $\triangle\text{ANC}$
$\angle\text{ANC}+\angle\text{NAC}+\angle\text{ACN}=180^\circ$ (Angle sum property)
$\angle\text{ANC}+45^\circ+20^\circ=180^\circ$ (AN is angle bisector of ∠A)
$\angle\text{ANC}=115^\circ$
In $\triangle\text{AMN}$
$\angle\text{AMN}+\angle\text{MAN}=\angle\text{ANC}$ (Measure of exterior angle is equla to the sum of two opposite interior angles)
$90^\circ+\angle\text{MAN}=115^\circ$
$\angle\text{MAN}=25^\circ.$

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MCQ 691 Mark

In figure, if $\frac{\text{y}}{\text{x}}=5$ and $\frac{\text{z}}{\text{x}}=4,$ then the value of $x$ is:

A
$8^\circ$
✓
$18^\circ$
C
$12^\circ$
D
$15^\circ$

Answer

Correct option: B.

$18^\circ$

From figure, we can see that
$\angle\text{x}^\circ+\angle\text{y}^\circ+\angle\text{z}^\circ=180^\circ\dots(1)$
Now,
$\frac{\text{y}}{\text{x}}=5\Rightarrow\ \text{y}=5\text{x}$
And,
$\frac{\text{z}}{\text{x}}=4,\text{z}=4\text{x}$
Substituting these value in equation (1), we have
$\angle\text{x}^\circ+\angle5\text{x}^\circ+\angle4\text{x}=180^\circ$
$\Rightarrow\ \angle10\text{x}^\circ=180^\circ$
$\Rightarrow\ \angle\text{x}^\circ=18^\circ$

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MCQ 701 Mark

In the given figure, $AB \| CD, CD \| EF$ and $y : z = 3 : 7$, then $x =?$

A
$108^\circ$
B
$162^\circ$
✓
$126^\circ$
D
$63^\circ$

Answer

Correct option: C.

$126^\circ$

$y : z = 3 : 7$
Let common ratio be a
$y = 3a$
$z = 7a$
$x = z$ (corresponding angle)
$x = 7a$
$x + y = 180^\circ $ (interior angle)
$7a + 3a = 180^\circ$
$10 a = 180^\circ$
$\text{a}=\frac{180}{10}$
$a = 18$
$x = 7a$
$x = 7 \times 18$
$x = 126^\circ .$

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MCQ 711 Mark

$AB$ and $CD$ are two parallel lines. $PQ$ cuts $AB$ and $CD$ at $E$ and $F$ respectively. $EL$ is the bisector of $\angle\text{FEB}.$ If $\angle\text{LEB}=35^\circ,$ then $\angle\text{CFQ}$ will be:

A
$130^\circ$
B
$70^\circ$
✓
$110^\circ$
D
$55^\circ$

Answer

Correct option: C.

$110^\circ$

It is given that, $AB \| CD$ with $PQ$ as transversal.
Also, $EL$ is the bisector $\angle\text{BEF}$ and $\angle\text{LEB}=35^\circ$
We need to find $\angle\text{CFQ}$
Therefore, $\angle\text{BEF}=2(\angle\text{LEB})$
$\angle\text{BEF}=2(35^\circ)$
$\angle\text{BEF}=70^\circ\text{ (i)}$
We have $AB \| CD$, $\angle\text{BEF}$ and $\angle\text{DEF}$ are consecutive interior angles, which must be supplementary.
$\angle\text{BEF}+\angle\text{DFE}=180^\circ$
From equation $(i)$, we get:
$70^\circ+\angle\text{DFE}=180^\circ$
$\angle\text{DFE}=180^\circ-70^\circ$
$\angle\text{DFE}=110^\circ\text{ (ii)}$
We have $\angle\text{CFQ}$ and $\angle\text{DFE}$ as vertically opposite angles.
Therefore,
$\angle\text{CFQ}=\angle\text{DFE}$
$\angle\text{CFQ}=110^\circ.$

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MCQ 721 Mark

In Fig., if line segment $AB$ is parallel to the line segment $CD$, what is the value of $y?$

A
$15$
B
$12$
✓
$20$
D
$18$

Answer

Correct option: C.

$20$

Since, $AB \| CD$ And, $BD$ cuts them
$y + 2y + y + 5y = 180^\circ $ (Consecutive interior angle)
$9y = 180^\circ $
$y = 20^\circ .$

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MCQ 731 Mark

In the adjoining figure, if $QP \| RT$, then $x$ is equal to:

✓
$75^\circ$
B
$70^\circ$
C
$65^\circ$
D
$55^\circ$

Answer

Correct option: A.

$75^\circ$

$\angle\text{QPR}=\angle\text{PRT}=40^\circ$ (Alternate interior angles)
In $\triangle\text{QPR}$
$\angle\text{PQR}+\angle\text{QPR}+\angle\text{PRQ}=180^\circ$(Angle sum property)
$65^\circ + 40^\circ + x^\circ = 180^\circ$
$x^\circ = 180^\circ - 40^\circ - 65^\circ$
$x = 75^\circ .$

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MCQ 741 Mark

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:

A
An isosceles triangle.
B
An equilateral triangle.
✓
A right triangle.
D
An obtuse angled triangle.

Answer

Correct option: C.

A right triangle.

The sum of the angles of triangle is $180$ degrees.
Let the angles of triangle be $a, b, c$
we have given that one angle of a triangle is equal to the sum of the other two angles
So we have,
$c = a + b$
$a+ b + c = 180$
Substitute c for $a + b$
$c + c = 180$
$2c = 180$
$c = 90$
Therefore the triangle is a right triangle.

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MCQ 751 Mark

In the adjoining figure, l || m then $\angle\text{x}$ is equal to:

✓
$56^\circ$
B
$51^\circ$
C
$66^\circ$
D
$61^\circ$

Answer

Correct option: A.

$56^\circ$

$x + 51^\circ = 107^\circ $ (Alternate interior angles)
$x = 107^\circ - 51^\circ = 56^\circ .$

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MCQ 761 Mark

In the given figure, sides $CB$ and $BA$ of $\triangle\text{ABC}$ have been produced to D and E respectively such that $\angle\text{ABD}=110^\circ$ and $\text{CAE}=135^\circ.$ Then $\angle\text{ACB}=?$

A
$45^\circ$
B
$55^\circ$
C
$35^\circ$
✓
$65^\circ$

Answer

Correct option: D.

$65^\circ$

$\angle\text{EAC}+\angle\text{BAC}=180^\circ$ (Linear Pair)
$\angle\text{EAC}=135^\circ$
$135^\circ+\angle\text{BAC}=180^\circ$
$\angle\text{BAC}=180^\circ-135^\circ$
$\angle\text{BAC}=45^\circ$
$\angle\text{ABD}+\angle\text{ABC}=180^\circ$(Linear Pair)
$\angle\text{ABD}=110^\circ$
$110^\circ+\angle\text{ABC}=180^\circ$
$\text{ABC}=180^\circ-110^\circ$
$\angle\text{ABC}=70^\circ$
In $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$
$45^\circ+70^\circ+\angle\text{ACB}=180^\circ$
$115^\circ+\angle\text{ACB}=180^\circ$
$\angle\text{ACB}=180^\circ-115^\circ$
$\angle\text{ACB}=65^\circ.$

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MCQ 771 Mark

In the given figure, $AB \| CD$. If $\angle\text{BAO}=60^\circ$ and $\angle\text{OCD}=110^\circ$ then $\angle\text{AOC}=?$

A
$70^\circ$
B
$60^\circ$
✓
$50^\circ$
D
$40^\circ$

Answer

Correct option: C.

$50^\circ$

Let $\angle\text{AOC}=\text{x}^\circ$
Draw $YOZ \| CD \| AB.$

Now, $YO\| AB$ and $OA$ is the transversal.
$\Rightarrow\angle\text{YOA}=\angle\text{OAB}=60^\circ$ (alternate angles)
Again, $OZ \| CD$ and $OC$ is the transversal.
$\Rightarrow\angle\text{COZ}+\angle\text{OCD}=180^\circ$ (interior angles)
$\Rightarrow\angle\text{COZ}+110^\circ=180^\circ$
$\Rightarrow\angle\text{COZ}=70^\circ$
Now, $\angle\text{YOZ}=180^\circ$ (straight angle)
$\Rightarrow\angle\text{YOA}+\angle\text{AOC}+\angle\text{COZ}=180^\circ$
$\Rightarrow60^\circ+\text{x}+70^\circ=1806^\circ$
$\Rightarrow\text{x}=50^\circ$
$\Rightarrow\angle\text{AOC}=50^\circ$

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MCQ 781 Mark

In Fig. if $AB \| CD$, then $x =$

A
$105^\circ$
B
$115^\circ$
✓
$100^\circ$
D
$110^\circ$

Answer

Correct option: C.

$100^\circ$

Given that,$AB \| CD$
Produce $P$ to $Q$ so that $PQ \| AB \| CD$
$\angle\text{BAP}+\angle\text{APQ}=180^\circ$ (Interior angle)
$132^\circ+\angle\text{APQ}=180^\circ$
$\angle\text{APQ}=48^\circ\text{(i)}$
$\angle\text{APC}=\angle\text{APQ}+\angle\text{QPC}$
$148^\circ=48^\circ+\angle\text{QPC}$ [From (i)]
$\angle\text{QPC}=100^\circ$
$\angle\text{QPC}+\angle\text{PCD}=180^\circ$ (Interior angles)
$100^\circ+\angle\text{PCD}=180^\circ$
$\angle\text{PCD}=80^\circ$
$\angle\text{PCD}+\text{x}=180^\circ$(Linear pair)
$80^\circ+\text{x}=180^\circ$
$\text{x}=100^\circ.$

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MCQ 791 Mark

The angles of a triangle are in the ratio $5 : 3 : 7$, the triangle is:

A
An isosceles triangle.
✓
An acute angled triangle.
C
An obtuse angled triangle.
D
A right triangle.

Answer

Correct option: B.

An acute angled triangle.

Let the angles of the triange be $5x, 3x$ and $7x$
We know that the sum of the angles of a triangle is $180^\circ $
$5x + 3x + 7x = 180^\circ$
$15x = 180^\circ$
$x = 12^\circ$
Therefore the angles are
$5x = 5 \times 120 = 60^\circ$
$3x = 3 \times 120 = 36^\circ$
$7x = 7 \times 120 = 84^\circ$
Since all the angles are less than $90^\circ $, therefore it is a acute angled triangle.

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MCQ 801 Mark

In the adjoining figure, $m \| n$. If $\angle\text{a}:\angle\text{b}=2:3,$ then the measure of $\angle\text{h}$ is:

A
$72^\circ$
✓
$108^\circ$
C
$150^\circ$
D
$120^\circ$

Answer

Correct option: B.

$108^\circ$

$\angle\text{a}+\angle\text{b}=180^\circ$ (Linear Pair)
$2\text{x}+3\text{x}=180^\circ$
$\text{x}=36^\circ$
$\angle\text{b}=36\times3=108^\circ$
$\angle\text{h}=\angle\text{b}=180^\circ$ (Alternate Exterior angle).

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MCQ 811 Mark

Given that lines $I_1, I_2$ and $I_3$ in figure are parallel. The value of $x$ is:

A
40$^{\circ}$
✓
140$^{\circ}$
C
80$^{\circ}$
D
50$^{\circ}$

Answer

Correct option: B.

140$^{\circ}$

In the given figure
$40^\circ+\angle\text{a} = 180^\circ$(linear - pair)
Therefore $\angle\text{a}=180^\circ-40^\circ=140^\circ$
Now $\angle\text{a}=\angle\text{b}$ (corresponding - angles)
Similarly $\angle\text{b}=\angle\text{x}=140^\circ$
Therefore $\angle\text{x}=140^\circ.$

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MCQ 821 Mark

In the adjoining figure, if $DE \| BC$, then the values of $x$ and $y$ are:

A
$x = 25^\circ , y = 85^\circ$
✓
$x = 25^\circ , y = 135^\circ$
C
$x = 260, y =1380$
D
$x = 20^\circ , y = 120^\circ$

Answer

Correct option: B.

$x = 25^\circ , y = 135^\circ$

$\angle\text{EAB}+\angle\text{ABC}=180^\circ$ (Sum of supplementary angle)
$2x - 5^\circ + 3x + 60^\circ = 180^\circ$
$5x = 180^\circ - 60^\circ + 5^\circ$
$x = 25^\circ$
$\angle\text{EAC}+\text{y}=180^\circ$ (Sum of supplementary angle).
$2x - 5^\circ + y = 180^\circ$
$y = 180^\circ + 5^\circ - 50^\circ = 135^\circ .$

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MCQ 831 Mark

An exterior angle of a triangle is $110^\circ $ and its two interior opposite angles are equal. Each of these equal angles is:

A
$70^\circ$
✓
$55^\circ$
C
$35^\circ$
D
$27\frac{1^\circ}{2}$

Answer

Correct option: B.

$55^\circ$

Let each interior opposite angle be $x.$
Then, $x + x = 110^\circ $ (Exterior angle property of a triangle)
$\Rightarrow 2x = 110^\circ$
$\Rightarrow x = 55^\circ$

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MCQ 841 Mark

In the adjoining figure $PQ \| XR$. If $\text{QP}\bot\text{SP},$ then the values of $x$ and $y$ are:

A
$x = 63^\circ , y = 27^\circ$
B
$x = 37^\circ , y =53^\circ$
C
$x = 43^\circ , y = 47^\circ$
✓
$x = 53^\circ , y = 37^\circ$

Answer

Correct option: D.

$x = 53^\circ , y = 37^\circ$

In $\triangle\text{QXR}$
$28^\circ+\angle\text{QXR}=\angle\text{QRT}$ (Exterior angle property)
$\text{QXR}=65^\circ-28^\circ=37^\circ$
$\text{y}=\angle\text{QSR}=37^\circ$ (Alternate interior angles)
In $\triangle\text{PQX}$
$\angle\text{PQX}+\angle\text{QXP}+\angle\text{QPX}=180^\circ$ (Angle sum property)
$37^\circ + x + 90^\circ = 180^\circ$
$x = 53^\circ .$

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MCQ 851 Mark

If two supplementary angles are in the ratio $2 : 7$, then the angles are:

A
$35^\circ , 145^\circ$
B
$70^\circ , 110^\circ$
✓
$40^\circ , 140^\circ$
D
$50^\circ , 130^\circ$

Answer

Correct option: C.

$40^\circ , 140^\circ$

We know that supplementary angles are those angles whose sum is $180^\circ $
The two given supplementary angles are in the ratio $2 : 7$
Let the commom ratio be $x$
So angles are $2x$ and $7x$ respectively
$2x + 7x = 180^\circ$
$9x = 180^\circ$
$\text{x}=\frac{180^\circ}{9}=20^\circ$
$2x = 2 \times 40^\circ = 40^\circ$
$7x = 7 \times 20^\circ = 140^\circ$

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MCQ 861 Mark

In Fig. if $l \| m$, what is the value of $x?$

A
$50$
B
$45$
✓
$60$
D
$30$

Answer

Correct option: C.

$60$

Given that,
$l \| m$
Let, $\angle1=3\text{y}$
$\angle2=2\text{y}+25^\circ$
$\angle3=\text{x}+15^\circ$
$\angle1=\angle2$ (Alternate angle)
$3y = 2y + 25^\circ$
$y = 25^\circ$
$\angle2=\angle3$ (Vertically opposite angle)
$x + 15^\circ = 2 (25^\circ ) + 25^\circ$
$x = 60^\circ .$

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MCQ 871 Mark

The sum of the angles of a triangle is:

✓
$180^\circ$
B
$0^\circ$
C
$360^\circ$
D
$90^\circ$

Answer

Correct option: A.

$180^\circ$

Line a and b are parallel.
$\angle\text{B'CA}=\angle\text{BAC} $ (Alternate interior angle)
$\angle\text{A'CB}=\angle\text{ABC}$ (Alternate interior angle)
$\angle\text{B'CA}+\angle\text{A'CB}+\angle\text{BCA}=180^\circ$(Linear Pair)
Therefore, $\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ.$

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MCQ 881 Mark

In the given figure $AB \| CD$ and $CD \| EF.$ If $y : z = 3 : 7$ then $x = ?$

A
$108^\circ$
✓
$126^\circ $
C
$162^\circ$
D
$63^\circ $

Answer

Correct option: B.

$126^\circ $

$AB \| CD$
$x + y = 180^\circ $ and $y = p$ (Vertically opposite angles)
Also, $CD \| EF$ and t is the transversal.
$\therefore p + z = 180^\circ $
$\Rightarrow y + z = 180^\circ $
$(\because\text{p}=\text{y})$
$\therefore x + y = y + z$
$\Rightarrow x = z$
But $y : z = 3 : 7$
$\therefore\text{y}=\Big(180\times\frac{3}{10}\Big)=54^\circ$ and $\text{z}=\Big(180\times\frac{7}{10}\Big)=126^\circ$
$x = 126^\circ $
$(\because\text{x}=\text{z})$

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MCQ 891 Mark

In the adjoining figure, $\text{m}\parallel\text{n}$ if $\angle1=50^\circ$ then $\angle2$ is equal to:

A
$50^\circ$
B
$40^\circ$
✓
$130^\circ$
D
$120^\circ$

Answer

Correct option: C.

$130^\circ$

$\angle2 = 180^\circ-\angle1$
$\angle2 = 180^\circ-50^\circ=130^\circ$

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MCQ 901 Mark

In figure, the value of $x$, is:

A
$12$
B
$15$
✓
$20$
D
$30$

Answer

Correct option: C.

$20$

From figure, we can see that
$\angle\text{BOD}+\angle\text{AOD}=180^\circ$
$\angle\text{BOD}=90^\circ$ [Given]
$\Rightarrow\ \angle\text{AOD}=180^\circ-90^\circ=90^\circ$
Now,
$\text{x}^\circ=\angle\text{COE}=\angle\text{FOD}$ [Opposite angles are equal]
Now,
$\angle\text{AOF}+\angle\text{FOD}=90^\circ=\angle\text{AOD}$
$\Rightarrow\ 3\text{x}^\circ+10^\circ+\text{x}^\circ=90^\circ$
$\Rightarrow\ 4\text{x}^\circ=80^\circ$
$\Rightarrow\ \text{x}^\circ=20^\circ$

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MCQ 911 Mark

The angles of a triangle are in the ration $3 : 5 : 7.$ The triangle is:

✓
Acute-angled.
B
Obtuse-angled.
C
Right-angled.
D
An isosceles triangle.

Answer

Correct option: A.

Acute-angled.

Let the angles measure $(3x)^\circ , (5x)^\circ $ and $(7x)^\circ .$
Then,
$3x + 5x + 7x = 180^\circ $
$\Rightarrow 15x = 180^\circ $
$\Rightarrow x = 12^\circ $
Therefore, the angles are $3(12)^\circ = 36^\circ , 5(12)^\circ = 60^\circ $ and $7(12)^\circ = 84^\circ .$
Hence, the triangle is acute-angled.

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MCQ 921 Mark

The difference between two complementary angles is $400$. The angles are:

A
$70^{\circ}, 110^{\circ}$
B
$65^{\circ}, 35^{\circ}$
✓
$65^{\circ}, 25^{\circ}$
D
$70^{\circ}, 30^{\circ}$

Answer

Correct option: C.

$65^{\circ}, 25^{\circ}$

We know that the sum of two complementary angles is $90^{\circ}$Let the two angles be $x$ and $y$
$x+y=90^{\circ}(1)$
We also know that the difference of the angles is $40^{\circ}$
Therefore, $x-y=40^{\circ}(2)$
Combining $(1)$ and $(2)$
We have
$x+y=90^{\circ}$
$x-y=40^{\circ}$
Solving as simultaneous equations we get
$2 x=130^{\circ}$
Hence $x=65^{\circ}$
Substituting this value of $x$ in either one of the equations $(1)$ or $(2)$
We get $y=25^{\circ}$.

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MCQ 931 Mark

An exterior angle of a triangle is 80$^{\circ}$ and the interior opposite angles are in the ratio $1 : 3$. Measure of each interior opposite angle is:

✓
$20^{\circ}, 60^{\circ}$
B
$40^{\circ}, 120^{\circ}$
C
$30^{\circ}, 90^{\circ}$
D
$30^{\circ}, 60^{\circ ^3}$

Answer

Correct option: A.

$20^{\circ}, 60^{\circ}$

Let the common ratio is $x$
The ratio of interior angles are $1 : 3$
So angles are $x$ and $3x$
$x + 3x = 80$
$4x = 80$
$\text{x}=\frac{80}{4}$
$x = 20$
So angles are $20^{\circ}$ and $60^{\circ}$.

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MCQ 941 Mark

In the adjoining figure, the bisectors of $\angle\text{CBD}$ and $\angle\text{BCE}$ meet at the point $O$. If $\angle\text{BAC}=70^\circ,$ then $\angle\text{BOC}$ is equal to:

A
$35^\circ$
B
$11^\circ$
✓
$55^\circ$
D
$70^\circ$

Answer

Correct option: C.

$55^\circ$

$\angle\text{BOC}=90^\circ-\frac{1}{2}\angle\text{BAC}$
$\angle\text{BOC}=90^\circ-35^\circ=55^\circ.$

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MCQ 951 Mark

If a transversal intersects two parallel lines, then which of the following alternatives is not true?

A
Each pair of corresponding angles is equal.
B
Each pair of alternate interior angles is equal.
✓
Each pair of co$-$interior angles is complementary.
D
Each pair of co$-$interior angles is supplementary.

Answer

Correct option: C.

Each pair of co$-$interior angles is complementary.

Co$-$interior angles are supplementary, not complementary.

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MCQ 961 Mark

In the given figure, $AB \| CD$. If $\angle\text{EAB}=50^\circ$ and $\angle\text{ECD}=60^\circ,$ then $\angle\text{AEB}=?$

✓
$70^\circ$
B
$55^\circ$
C
$60^\circ$
D
$50^\circ$

Answer

Correct option: A.

$70^\circ$

$\angle\text{BCD}=\angle\text{ABE}=60^\circ$ (Vertically opposite angle)
In $\triangle\text{EAB}$
$\angle\text{EAB}+\angle\text{EBA}+\angle\text{AEB}+180^\circ$ (Angle sum property)
$50^\circ+60^\circ+\angle\text{AEB}=180^\circ$
$\text{AEB}=70^\circ.$

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MCQ 971 Mark

In figure, the value of $y$ is:

A
$20^\circ$
✓
$30^\circ$
C
$45^\circ$
D
$60^\circ$

Answer

Correct option: B.

$30^\circ$

From figure, we can see
$\angle\text{x}^\circ=\angle\text{y}^\circ$ [Vertically opposite angles]
Also,
$\angle\text{3}\text{x}^\circ+\angle\text{y}^\circ+\angle2\text{x}^\circ=180^\circ$
Now,
$\angle\text{x}^\circ=\angle\text{y}^\circ$
$\therefore\ \angle\text{3}\text{y}^\circ+\angle\text{y}^\circ+\angle2\text{y}^\circ=180^\circ$
$\Rightarrow\ \angle6\text{y}^\circ=180^\circ$
$\Rightarrow\ \angle\text{y}^\circ=180^\circ$

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MCQ 981 Mark

If two lines intersect each other then:

✓
Vertically opposite angles are equal.
B
Corresponding angles are equal.
C
Co-interior angles are equal.
D
Alternate interior angles are equal.

Answer

Correct option: A.

Vertically opposite angles are equal.

$\angle\text{A}+\angle\text{B}=180 ($Linear Pair$)$
$\angle\text{B}+\angle\text{C}=180 ($Linear Pair$)$
On equating ablove equations, we get
$\angle\text{A}+\angle\text{B}=\angle\text{B}+\angle\text{C}$
$\angle\text{A}=\angle\text{C}$
Similarly, $\angle\text{B}=\angle\text{D}.$

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MCQ 991 Mark

In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=(3\text{x}+10)^\circ$ and $\angle\text{BOC}=(4\text{x}−26)^\circ.$ then $\angle\text{BOC}=?$

A
$76^\circ$
B
$106^\circ$
C
$96^\circ$
✓
$86^\circ$

Answer

Correct option: D.

$86^\circ$

We have,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$ [Since $AOB$ is a straight line]
$\Rightarrow 3x + 10 + 4x - 26 = 180^\circ$
$\Rightarrow 7x = 196^\circ$
$\Rightarrow x = 28^\circ$
$\therefore\angle\text{BOC}=[4\times28−26]^\circ$
Hence, $\angle\text{BOC}=86^\circ.$

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MCQ 1001 Mark

Side BC of $\triangle\text{ABC}$ has been produced to $D$ on left-hand side and to $E$ on right-hand side such that $\angle\text{ABD}=125^\circ$ and $\angle\text{ACE}=130^\circ.$ Then $\angle\text{A}=?$

A
$50^\circ$
✓
$75^\circ$
C
$55^\circ$
D
$65^\circ$

Answer

Correct option: B.

$75^\circ$

$\angle\text{ABD}+\angle\text{ABC}=180^\circ$ (Linear Pair)
$\angle\text{ABC}=180^\circ-125^\circ=55^\circ$
$\angle\text{ACE}+\angle\text{ACB}=180^\circ$ (Linear Pair)
$\angle\text{ACB}=180^\circ-130^\circ=50^\circ$
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$ (Angle sum property)
$\angle\text{BAC}=180^\circ-50^\circ-55^\circ$
$\angle\text{BAC}=75^\circ.$

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MCQ 1011 Mark

In two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $5 : 4$, then the smaller of the two angles is:

A
$120^\circ$
B
$60^\circ$
C
$100^\circ$
✓
$80^\circ$

Answer

Correct option: D.

$80^\circ$

We know that sum of two interior angles on the same side of atransversal intersecting two parallel lines is $180^\circ$
let the common ratio is $x$ so the angles are $5x ,4x$
$So 5x + 4x = 180^\circ$
$9x = 180^\circ$
$\text{x}=\frac{180^\circ}{9}$
$\text{x}=20^\circ$
So the angles are $5x = 100^\circ$
$4x = 80^\circ$
So smallest angle is $80^\circ$

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MCQ 1021 Mark

An angle is one-fifth of its supplement. The measure of the angle is:

A
$75^\circ$
B
$15^\circ$
✓
$30^\circ$
D
$15^\circ$

Answer

Correct option: C.

$30^\circ$

Let one angle be$ x^\circ $
Its supplementary angle will be $180^\circ - x^\circ $
According to question
$\text{x}=\frac{1}{5}(180^\circ-\text{x})$
$5x + x = 180^\circ $
$6x = 180^\circ $
$\text{x}=\frac{180}{6}$
$x = 30^\circ .$

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MCQ 1031 Mark

The measure of the Complementary angle of $63^{\circ}$ is:

✓
$27^{\circ}$
B
$117^{\circ}$
C
$30^{\circ}$
D
$36^{\circ}$

Answer

Correct option: A.

$27^{\circ}$

Sum of complementary angles is 90$^{\circ}$.If one angle is 63$^{\circ}$
Then the other angle will be 90$^{\circ}$ - 63$^{\circ}$= 27$^{\circ}$.

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MCQ 1041 Mark

In figure, if lines l and m are parallel lines, then $x =$

A
$70^\circ$
B
$100^\circ$
✓
$40^\circ$
D
$30^\circ $

Answer

Correct option: C.

$40^\circ$

From figure,
$\angle\text{ABC}=\angle\text{DCE}\dots(1)$ [Corresponding angles]
$\angle\text{ECF}=180^\circ-\angle\text{DCE}$ [Supplementary]
$=180^\circ-\angle\text{ABC}$ [From (1)]
$=180^\circ-70^\circ$
$\Rightarrow\ \angle\text{ECF}=110^\circ$
Now, in $\triangle\text{CEF}$
$\angle\text{ECF}+\angle\text{CFE}+\angle\text{FEC}=180^\circ$
$\Rightarrow\ 110^\circ+\text{x}+30^\circ=180^\circ$
$\Rightarrow\ \text{x}=40^\circ$

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MCQ 1051 Mark

Write the correct answer in the following: Angles of a triangle are in the ratio $2 : 4 : 3$. The smallest angle of the triangle is,

A
$60^\circ$
✓
$40^\circ$
C
$80^\circ$
D
$20^\circ$

Answer

Correct option: B.

$40^\circ$

Given that: The Ratio of angles of a triangle is $2 : 4 : 3$
Let the angles of the triangle be $\angle\text{A},\angle\text{B},$ and $\angle\text{C},$
$\therefore\ \angle\text{A}=2\text{x},\angle\text{B}=4\text{x}$ and $\angle\text{C}=3\text{x}$
In $\angle\text{ABC},$ $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
[$\because\ $Sum of angles of a triangle is $180^\circ ]$
$\Rightarrow2\text{x}+4\text{x}+3\text{x}=180^\circ\Rightarrow9\text{x}=180^\circ\Rightarrow\text{x}=\frac{180^\circ}{9}=20^\circ$
$\therefore\ \angle\text{A}=2\text{x}=2\times20^\circ=40^\circ$
$\angle\text{B}=4\text{x}=4\times20^\circ=80^\circ$
And $\angle\text{C}=3\text{x}=3\times20^\circ=60^\circ$
Hence, the smallest angles of a triangle is $40^\circ $ and option $(b)$ is correct answer.

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MCQ 1061 Mark

Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is:

A
$45^\circ$
✓
$36^\circ$
C
$30^\circ$
D
None of these

Answer

Correct option: B.

$36^\circ$

Let x and $(90^\circ - x)$ be two complimentary angles
According to question,
$2x = 3 (90^\circ - x)$
$2x = 270^\circ - 3x$
$x = 54^\circ $
The angles are:
$54^\circ $ and $90^\circ - 54^\circ = 36^\circ$
Thus, smallest angle is $36^\circ .$

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MCQ 1071 Mark

An exterior angle of a triangle is $80^\circ $ and two interior opposite angles are equal. Measure of each of these angles is:

A
$100^\circ$
B
$120^\circ$
✓
$40^\circ$
D
$60^\circ$

Answer

Correct option: C.

$40^\circ$

We know that the exterior angle so formed is equal to the sum of the two interior opposite angles.Let the two interior opposite angles be $x.$
So, $x + x = 80^\circ$
$2x = 80^\circ$
$x = 40^\circ .$

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MCQ 1081 Mark

In figure, if $CP \| BQ$, then the measure of $x$ is:

✓
$130^\circ$
B
$105^\circ$
C
$175^\circ$
D
$125^\circ$

Answer

Correct option: A.

$130^\circ$

From figure,
$\angle\text{QBA}=\angle\text{CEA}$ [Correspondence angles]
$\Rightarrow\ \angle\text{CEA}=105^\circ\dots(1)$
In $\triangle\text{ACE},$
$\angle\text{CEA}+\angle\text{EAC}+\angle\text{ACE}=180^\circ$
$\Rightarrow\ 105^\circ+25^\circ+\angle\text{ACE}=180^\circ$ [From (1)]
$\Rightarrow\ 130^\circ+\angle\text{ACE}=180^\circ$
$\Rightarrow\ \angle\text{ACE}=50^\circ$
Now,
$\text{x}=\angle\text{ACP}=180^\circ-\angle\text{ACE}$
$=180^\circ-50^\circ=130^\circ$

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MCQ 1091 Mark

$AB$ and $CD$ are two parallel lines. $PQ$ cuts $AB$ and $CD$ at $E$ and $F$ respectively. $EL$ is the bisector of $\angle\text{FEB}.$ If $\angle\text{LEB}=35^\circ,$ then $\angle\text{CFQ}$ will be:

✓
$55^\circ$
B
$70^\circ$
C
$110^\circ$
D
$130^\circ$

Answer

Correct option: A.

$55^\circ$

From figure,
$\angle\text{LEB}=\angle\text{FEL}$ [EL is bisector of $\angle\text{FEB}$]
Now,
$\angle\text{FEB}=2\angle\text{LEB}=2\times35^\circ=70^\circ$
Also,
$\angle\text{FEB}=\angle\text{CFE}$ [Alternate interior angles]
$\Rightarrow\ \angle\text{CFE}=70^\circ$
Now,
$\angle\text{CFE}+\angle\text{CFQ}=180^\circ$
$\Rightarrow\ 70^\circ+\angle\text{CFQ}=180^\circ$
$\Rightarrow\ \angle\text{CFQ}=180^\circ-70^\circ=110^\circ$

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MCQ 1101 Mark

In Fig. if $l1 \| l2$, what is the value of $y?$

A
$120$
✓
$135$
C
$100$
D
$150$

Answer

Correct option: B.

$135$

Given that,$l1 \| l2$ and $l3$ is transversal
$\angle1=3\text{x}$ (Vertically opposite angle)
$\text{y}=\angle1$ (Corresponding angle)
$y=3 x \text { (i) }$
$y+x=180^{\circ} \text { (Linear pair) }$
$3 x+x=180^{\circ}[\text { From (i) }]$
$4 x=180^{\circ}$
$x=45^{\circ}$
Therefore,
$y=3 x=3 \times 45^{\circ}$
$=135^{\circ}$.

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MCQ 1111 Mark

If two supplementary angles are in the ratio $2 : 7$, then the angles are:

✓
$40^\circ , 140^\circ$
B
$50^\circ , 130^\circ$
C
$70^\circ , 110^\circ$
D
$35^\circ , 145^\circ$

Answer

Correct option: A.

$40^\circ , 140^\circ$

We know that supplementary angles are those angles whose sum is $180^\circ $
The two given supplementary angles are in the ratio $2 : 7$
Let the commom ratio be $x,$
So angles are $2x$ and $7x$ respectively $2x + 7x = 180^\circ$
$\text{x}=\frac{180^\circ}{9}=20^\circ$
$2x = 2 \times 40 = 40^\circ$
$7x = 7 \times 20^\circ = 140^\circ .$

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MCQ 1121 Mark

If two angles are complements of each other, then each angle is:

✓
An acute angle.
B
An obtuse angle.
C
A right angle.
D
A reflex angle.

Answer

Correct option: A.

An acute angle.

If two angles are complements of each other, that is, the sum of their measures is $90^\circ $, then each angle is an acute angle.

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MCQ 1131 Mark

In the given figure, $AB \| CD$, If $\angle\text{ABO}=45^\circ$ and $\angle\text{COD}=100^\circ$ then $\angle\text{CDO}=?$

A
$45^\circ$
✓
$35^\circ$
C
$25^\circ$
D
$30^\circ$

Answer

Correct option: B.

$35^\circ$

$\angle\text{ABC}=\angle\text{BCD}=45^\circ$ (Alternate interior angles)
In $\triangle\text{COD}$
$\angle\text{COD}+\angle\text{CDO}+\angle\text{DCO}=180^\circ$ (Angle sum property)
$\angle\text{CDO}=180^\circ-100^\circ-45^\circ$
$\angle\text{CDO}=35^\circ.$

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MCQ 1141 Mark

In Fig. if 1$_1 \| 1_2$, what is $x + y $ in terms of w and $z?$

✓
$180 - w + z$
B
$180 + w + z$
C
$180 + w - z$
D
$180 - w - z$

Answer

Correct option: A.

$180 - w + z$

Given that,
$1_1 \| 1_2$
Let m and n be two transversal cutting them
$\angle\text{w}+\angle\text{x}=180^\circ$ (Consecutive interior angle)
$x=180^{\circ}-w(i)$
$z = y (Alternate angles) (ii)$
From (i) and (ii), we get
$x+y=180^{\circ}-w+z$.

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MCQ 1151 Mark

The complement of $(90^{\circ} - a)$ is:

✓
$a^{\circ}$
B
$-a^{\circ}$
C
$90^{\circ}$+ a
D
$90^{\circ}$ - a

Answer

Correct option: A.

$a^{\circ}$

Two angles, whose sum is$ 90^{\circ}$, are called the complementary angle.Let x is a complimentary angle of $(90^{\circ} - a)$
$x+\left(90^{\circ}-a\right)=90^{\circ}$
$x=90^{\circ}-\left(90^{\circ}-a\right)$
$x=90^{\circ}-90^{\circ}+a$
$x=a^{\circ} .$

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MCQ 1161 Mark

In the adjoining figure, $AOB$ is a straight line. If $x : y : z = 4 : 5 : 6$, then $y =?$

✓
$60^\circ$
B
$80^\circ$
C
$48^\circ$
D
$72^\circ$

Answer

Correct option: A.

$60^\circ$

The ratio of the angles is given to be $4 : 5 : 6.$
So, let the measure of the angles be $4\ m, 5\ m$ and $6\ m$.
Since $AOB$ is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow4\text{m}+5\text{m}+6\text{m}=180^\circ$
$\Rightarrow15\text{m}=180$
$\Rightarrow\text{m}=12$
So, $y = 5m = 5(12) = 60^\circ .$

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MCQ 1171 Mark

In the given figure, $AOB$ is a straight line. The value of $x$ is:

A
$12$
✓
$15$
C
$20$
D
$25$

Answer

Correct option: B.

$15$

$AOB$ is a straight line.
$\Rightarrow\angle\text{AOB}=180^\circ$
$\Rightarrow 60^\circ + 5x^\circ + 3x^\circ = 180^\circ$
$\Rightarrow 60^\circ + 8x^\circ = 180^\circ$
$\Rightarrow 8x^\circ = 120^\circ$
$\Rightarrow x = 15^\circ$

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MCQ 1181 Mark

Write the correct answer in the following: In Fig. $POQ$ is a line.The value of $x$ is.

✓
$20^\circ$
B
$25^\circ$
C
$30^\circ$
D
$35^\circ$

Answer

Correct option: A.

$20^\circ$

We have $3x + 4x + 40^\circ = 180^\circ $ (Angles on the straight line)
$\Rightarrow 7x + 40^\circ = 180^\circ $
$\Rightarrow 7x = 180^\circ - 40^\circ = 140^\circ $
$\Rightarrow x = 140^\circ ÷ 7 = 20^\circ $
Hence, $(a)$ is the correct answer.

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MCQ 1191 Mark

Write the correct answer in the following: An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is,

A
$37\frac{1}{2}^\circ$
✓
$52\frac{1}{2}^\circ$
C
$72\frac{1}{2}^\circ$
D
$75^\circ$

Answer

Correct option: B.

$52\frac{1}{2}^\circ$

An exterior angle of triangle is $150^\circ $
Let each of to two interior opposites angles be x.
We know that exterior angle of a equal to the sum of two interior opposite angles.
$105^\circ=\text{x}+\text{x}\Rightarrow2\text{x}=105^\circ$
$\Rightarrow\text{x}=\frac{1}{2}\times105^\circ=52\frac{1}{2}^\circ$
So, each of equal angle angle is $52\frac{1}{2}^\circ$
Hence, $(b)$ is the correct answer.

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MCQ 1201 Mark

In the adjoining figure, the value of $x$ is:

A
$18^\circ$
✓
$15^\circ$
C
$12^\circ$
D
$10^\circ$

Answer

Correct option: B.

$15^\circ$

$7x + 2x + 3x = 180^\circ$
$12x = 180^\circ$
$x = 15^\circ .$

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MCQ 1211 Mark

The incorrect statement is:

A
Two lines drawn in a plane always intersect at a point.
B
A line segment has definite length.
C
Three lines are concurrent if and only if they have a common point.
✓
One and only one line can be drawn passing through a given point and parallel to a given line.

Answer

Correct option: D.

One and only one line can be drawn passing through a given point and parallel to a given line.

If two lines intersect then they must lie in one plane but its converse is not necessarily true.

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MCQ 1221 Mark

Two straight lines $AB$ and $CD$ intersect one another at the point $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ, $ then $\angle\text{AOD}=$

✓
$86^{\circ}$
B
$137^{\circ}$
C
$90^{\circ}$
D
$94^{\circ}$

Answer

Correct option: A.

$86^{\circ}$

Given,
$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ\text{ (i)}$
$\angle\text{AOD}+\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=360^\circ$(Angles at a point)
$\angle\text{AOD}+274^\circ=360^\circ$
$\angle\text{AOD}=86^\circ.$

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MCQ 1231 Mark

If two angles of a triangle are $30^\circ $ and $45^\circ $, what is the measure of the third angle?

✓
$105^\circ$
B
$95^\circ$
C
$90^\circ$
D
$60^\circ$

Answer

Correct option: A.

$105^\circ$

We know that the sum of all angles of a triangle is $180^\circ $
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
Let $\angle\text{A}=30^\circ,\angle\text{B}=45^\circ$
$30^\circ+45^\circ+\angle\text{C}=180^\circ$
$75^\circ+\angle\text{C}=180^\circ$
$\angle\text{C}=180^\circ-75^\circ$
$\angle\text{C}=105^\circ.$

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MCQ 1241 Mark

In Fig., if lines $l$ and $m$ are parallel, then the value of $x$ is:

A
$75^{\circ}$
B
$65^{\circ}$
C
$55^{\circ}$
✓
$35^{\circ}$

Answer

Correct option: D.

$35^{\circ}$

Given that,
$l \| m$ and n cuts them
Let, $\angle1=\text{x}$
$\angle2=90^\circ$
$\angle3=125^\circ$
$\angle3+\angle5=180^\circ$(Linear pair)
$125^\circ+\angle5=180^\circ$
$\angle5=55^\circ\text{ (i)}$
$\angle4=99^\circ\text{ (ii)}$
Now,
$\angle1+\angle4+\angle5=180^\circ$ (Angle sum property)
$\text{x}+90^\circ+55^\circ=180^\circ$
$\text{x}=35^\circ.$

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MCQ 1251 Mark

Two angles measure $(70 + 2x)^\circ $ and $(3x - 15)^\circ $. If each angle is the supplement of the other, then the value of $x$ is:

✓
$25$
B
$250^\circ $
C
$30$
D
$20$

Answer

Correct option: A.

$25$

$70 + 2x + 3x - 15 = 180$ (Supplementary angles)
$5x = 180 - 55$
$x = 25.$

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MCQ 1261 Mark

In figure, if lines $l$ and $m$ are parallel, then $x =$

A
$20^\circ$
✓
$45^\circ$
C
$65^\circ$
D
$85^\circ$

Answer

Correct option: B.

$45^\circ$

From figure,
$\angle\text{ABD}=\angle\text{CDF}$ [Correspondence angles]
$\Rightarrow\ \angle\text{CDF}=65^\circ$
Now,
$\angle\text{FDE}=180^\circ-\angle\text{CDF}=180^\circ-65^\circ$
$\Rightarrow\ \angle\text{FDE}=115^\circ$
In $\triangle\text{EDF},$
$\angle\text{FDE}+\angle\text{DEF}+\angle\text{EFD}=180^\circ$
$\Rightarrow\ 115^\circ+\text{x}+20^\circ=180^\circ$ [Sum of all interior angles of a $\triangle$ as 180°]
$\Rightarrow\ \text{x}=180^\circ-20^\circ-115^\circ=45^\circ$

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MCQ 1271 Mark

In figure, if $I_1 \| I_2$, what isb the value of $y?$

A
$100$
B
$120$
✓
$135$
D
$150$

Answer

Correct option: C.

$135$

Let angle supplement of $3 x^{\circ}$ be $Z^{\circ}$
$\Rightarrow z^{\circ}=180^{\circ}-3 x^{\circ}$
$\Rightarrow\ \angle\text{AHF}+\angle\text{FHB}=180^\circ$
$\Rightarrow z^{\circ}+3 x^{\circ}=180^{\circ}$
$\Rightarrow z^{\circ}=180^{\circ}-3 x^{\circ}$
Now,
$x^{\circ}+y^{\circ}=180^{\circ}$
Also,
$x^{\circ}=z^{\circ} \text { [Correspondence angles] }$
$\Rightarrow x^{\circ}=180^{\circ}-3 x^{\circ}$
$\Rightarrow 4 x^{\circ}=180^{\circ}$
$\Rightarrow x^{\circ}=45^{\circ}$
$x^{\circ}+y^{\circ}=180^{\circ}$
$\Rightarrow y^{\circ}=180^{\circ}-x^{\circ}=180^{\circ}-45^{\circ}=135^{\circ}$.

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MCQ 1281 Mark

An exterior angle of a triangle is $80^\circ $ and the interior opposite angles are in the ratio $1 : 3$. Measure of each inte4rior opposite angle is:

A
$30^\circ , 60^\circ$
✓
$20^\circ , 60^\circ$
C
$30^\circ , 90^\circ$
D
$40^\circ , 120^\circ$

Answer

Correct option: B.

$20^\circ , 60^\circ$

let the common ratio is $x$
the ratio of interior angles are $1 : 3$
so angles are $x$ and $3x$
$x + 3x = 80$
$\text{x}=\frac{80}{4}$
$x = 20$
So angles are $20^\circ $ and $60^\circ $

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MCQ 1291 Mark

In Fig. if $AB \| HF$ and $DE \| FG$, then the measure of $\angle\text{FDE}$ is:

✓
$80^\circ$
B
$90^\circ$
C
$108^\circ$
D
$100^\circ$

Answer

Correct option: A.

$80^\circ$

Given that,
$AB \| HF$ and $CD$ cuts them
$\angle\text{HFC}=\angle\text{FDA}$ (Corresponding angle)
$\angle\text{FDA}=28^\circ$
$\angle\text{FDA}+\angle\text{FDE}+\angle\text{EDB}=180^\circ $(Linear pair)
$28^\circ+\angle\text{FDE}+72^\circ=180^\circ$
$\angle\text{FDE}=80^\circ.$

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MCQ 1301 Mark

In the adjoining figure, if $AB \| DE$, then the measure of $\angle\text{ACD}$ is:

A
$70^\circ$
B
$80^\circ$
C
$90^\circ$
✓
$100^\circ$

Answer

Correct option: D.

$100^\circ$

$x + 110^\circ = 180^\circ $(Supplementary angles)
$x = 70^\circ $
$y + 150^\circ = 180^\circ $(Supplementary angles)
$y = 30^\circ $
$\angle\text{ACD}=70^\circ+30^\circ=100^\circ.$

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MCQ 1311 Mark

The sum of all the angles of a quadrilateral is:

A
$400^\circ$
B
$180^\circ$
✓
$360^\circ$
D
$320^\circ$

Answer

Correct option: C.

$360^\circ$

Sum of the angles of a polygon $= (n - 2) \times 180^\circ $
Quadrilateral has $4$ sides,
So sum of interior angles $= (4 - 2) \times 180^\circ = 360^\circ .$

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MCQ 1321 Mark

In the given figure, $AB \| CD$, If $\angle\text{CAB}=80^\circ$ and $\angle\text{EFC}=25^\circ$ then $\angle\text{CEF}=?$

✓
$55^\circ $
B
$75^\circ $
C
$65^\circ$
D
$45^\circ$

Answer

Correct option: A.

$55^\circ $

$\angle\text{BAF}=\angle\text{DCF}=80^\circ$ (Corresponding angle)
In $\triangle\text{CEF}$
$\angle\text{CFE}+\angle\text{CEF}=\angle\text{DCF}$ (exterior angle is equal to the sum of its two opposite interior angles)
$25^\circ+\angle\text{CEF}=80^\circ$
$\angle\text{CEF}=55^\circ$

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MCQ 1331 Mark

In the adjoining figure, $BE$ and $CE$ are bisectors of $\angle\text{ABC}$ and $\angle\text{ACD}$ respectively. If $\angle\text{BEC}=25^\circ$ then $\angle\text{BAC}$ is equal to:

✓
$50^\circ$
B
$25\frac{1}{2}^\circ$
C
$12\frac{1}{2}^\circ$
D
$65^\circ$

Answer

Correct option: A.

$50^\circ$

$\angle\text{BEC}+\angle\text{EBC}=\angle\text{ECD} ($Exterior angle property$)$
$\angle\text{BEC}=\angle\text{ECD}-\angle\text{EBC}$
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}=\angle\text{ACD}$
$\angle\text{ABC}+2\angle\text{EBC}=2\angle\text{ECD}$
$\angle\text{ABC}=2(\angle\text{ECD}-\angle\text{EBC})$
$\angle\text{ABC}=2(\angle\text{BEC})$
$\angle\text{ABC}=50^\circ.$

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MCQ 1341 Mark

✓
$27^\circ , 121^\circ$
B
$20^\circ , 80^\circ$
C
$26^\circ , 124^\circ$
D
$121^\circ , 20^\circ$

Answer

Correct option: A.

$27^\circ , 121^\circ$

In $\triangle\text{PQR}$
$\angle\text{QPR}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$ (Angle sum property)
$\angle\text{PQR}=180^\circ-62^\circ-64^\circ$
$\angle\text{PQR}=54^\circ$
$\angle\text{ORQ}=32^\circ$ (OR is a bisector)
$\angle\text{OQR}=27^\circ$ OQ is a bisector)
In $\triangle\text{OQR}$
$\angle\text{OQR}+\angle\text{ORQ}+\angle\text{QOR}=180^\circ$ (Angle sum property)
$\angle\text{QOR}=180^\circ-32^\circ-27^\circ=121^\circ.$

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MCQ 1351 Mark

Two straight lines $AB$ and $CD$ intersect one another at the point $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ,$ then $\angle\text{AOD}=$

✓
$86^\circ$
B
$90^\circ$
C
$94^\circ$
D
$137^\circ$

Answer

Correct option: A.

$86^\circ$

$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}+\angle\text{AOD}=360^\circ\dots(1)$
Now,
$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ\dots(2)$ [Given]
From (1) and (2).
$274^\circ+\angle\text{AOD}=360^\circ$
$\Rightarrow\ \angle\text{AOD}=360^\circ-274^\circ$
$\Rightarrow\ \angle\text{AOD}=86^\circ$

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MCQ 1361 Mark

In the adjoining figure, $m \| n$, if $\angle1=50^\circ$ then $\angle2$ is equal to:

✓
$130^\circ$
B
$120^\circ$
C
$40^\circ$
D
$50^\circ$

Answer

Correct option: A.

$130^\circ$

$\angle2=180^\circ-\angle1$
$\angle2=180^\circ-50^\circ=130^\circ.$

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MCQ 1371 Mark

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $2 : 3$, then the measure of the larger angle is:

A
$54^\circ$
B
$120^\circ$
✓
$108^\circ$
D
$136^\circ$

Answer

Correct option: C.

$108^\circ$

Let $AB$ and $CD$ are two parallel lines and $PQ$ is transverce to it.
According to question,
$\frac{\angle\text{BRS}}{\angle\text{DSR}}=\frac{2}{3}$
$\Rightarrow\ \angle\text{BRS}=\frac{2}{3}\angle\text{DSR}\dots(1)$
Now,
$\angle\text{CSR}=\angle\text{BRS}$ [Alternate angles]
$\Rightarrow\ \angle\text{CSR}+\angle\text{DSR}=180^\circ$
$\Rightarrow\ \angle\text{BRS}+\angle\text{DSR}=180^\circ$
$\Rightarrow\ \frac{2}{3}\angle\text{DSR}+\angle\text{DSR}=180^\circ$
$\Rightarrow\ \angle\text{DSR}=\frac{180\times3}{5}=108^\circ$
$\Rightarrow\ \angle\text{BRS}=\frac{2}{3}\times108^\circ=72^\circ$
Thus,
$\angle\text{DSR}=108^\circ$ and $\angle\text{BRS}=72^\circ$
$\Rightarrow$ Larger angle is $\angle\text{DSR}.$

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MCQ 1381 Mark

The sum of all the angles of a quadrilateral is:

A
$180^\circ$
✓
$360^\circ$
C
$400^\circ$
D
$320^\circ$

Answer

Correct option: B.

$360^\circ$

Sum of the angles of a polygon $= (n - 2) \times 180$
Quadrilateral has $4$ sides,
So sum of interior angles $= (4 - 2) \times 180^\circ = 360$

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MCQ 1391 Mark

In figure,$ AOB$ is a straight line. If $\angle\text{AOC}+\angle\text{BOD}=85^\circ,$ then $\angle\text{COD}=$

A
$85^\circ $
B
$90^\circ$
✓
$95^\circ$
D
$100^\circ$

Answer

Correct option: C.

$95^\circ$

From figure, we can see
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
But,
$\angle\text{AOC}+\angle\text{BOD}=85^\circ$ [Given]
$\Rightarrow\ 85^\circ+\angle\text{COD}=180^\circ$
$\Rightarrow\ \angle\text{COD}=95^\circ$

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MCQ 1401 Mark

In figure, if $I_1 \| I_2$ and $I_3 \| I_4$, what is $y$ in terms of $x?$

A
$90+\text{x}$
B
$90+2\text{x}$
✓
$90-\frac{\text{x}}{2}$
D
$90-2\text{x}$

Answer

Correct option: C.

$90-\frac{\text{x}}{2}$

From figure,
$\angle\text{EPR}=\angle\text{PQS}$ [Correspondence angles are equal]
$\Rightarrow\ \angle\text{PQS}=\text{x}^\circ$
Also,
$\angle\text{PQS}=\angle\text{RSD}$ [Correspondence angles are equal]
$\Rightarrow\ \angle\text{RSD}=\text{x}^\circ$
Now,
$\angle\text{RSD}+\text{y}^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow\ \text{x}^\circ+2\text{y}^\circ=180^\circ$
$\Rightarrow\ \text{y}^\circ=\frac{180^\circ-\text{x}^\circ}{2}$
$\Rightarrow\ \text{y}^\circ=90^\circ-\frac{\text{x}^\circ}{2}$

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MCQ 1411 Mark

The number of triangles that can be drawn having angles as $50^\circ , 60^\circ $ and $70^\circ $ are:

A
None of these
B
Two
C
Only one
✓
Infinite

Answer

Correct option: D.

Infinite

As we know similar triangles can be drawn for any given triangle.
These similar triangles will have the same angles as the original triangle
$(\text{i.e}\angle50^\circ,\angle60^\circ$ and $\angle70^\circ)$ and will be infinite in number.

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MCQ 1421 Mark

In the given figure, $\angle\text{OAB}=110^\circ$ and $\angle\text{BCD}=130^\circ$ then $\angle\text{ABC}$ is equal to:

A
$40^\circ$
B
$50^\circ$
✓
$60^\circ$
D
$70^\circ$

Answer

Correct option: C.

$60^\circ$

Through $B$ draw $YBZ \| OA \| CD.$

Now, $OA \| YB$ and $AB$ is the transversal.
$\Rightarrow\angle\text{OAB}+\angle\text{YBA}=180^\circ$ (interior angles are supplementary)
$\Rightarrow\angle\text{YBA}=70^\circ$
Also, $CD \| BZ$ and $BC$ is the transversal.
$\Rightarrow\angle\text{DCB}+\angle\text{CBZ}=180^\circ$ (interior angles are supplementary)

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MCQ 1431 Mark

In the given figure, $AB \| CD$. If $\angle\text{CAB}=180^\circ$ and $\angle\text{EFC}=25^\circ$ then $\angle\text{CEF}=?$

A
$65^\circ$
✓
$55^\circ$
C
$45^\circ$
D
$75^\circ$

Answer

Correct option: B.

$55^\circ$

Since $AB \| CD$,$\Rightarrow\angle\text{ACE}=\angle\text{BAC}=80^\circ$
In $\triangle\text{CEF},$
$\angle\text{ACE}=\angle\text{CEF}+\angle\text{CFE}$ (Exterior angle is equal to sum of the remote interior angles)
$\Rightarrow80^\circ=\angle\text{CEF}+25^\circ$
$\Rightarrow\angle\text{CEF}=55^\circ$

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MCQ 1441 Mark

How many triangles can be drawn having angles as $45^\circ , 60^\circ $ and $85^\circ ?$

A
Two.
B
Infinitely many.
✓
None.
D
Only one.

Answer

Correct option: C.

None.

If we add up the three given angles we get
$45^\circ + 60^\circ + 85^\circ = 190^\circ$
But as we know that the angles of any triangle only add-up to $180^\circ .$
Hence the above-given triangle is not possible.

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MCQ 1451 Mark

When two straight lines intersect:
$i$. Adjacent angles are complementary
$ii$. Adjacent angles are supplementary.
$iii$. Opposite angles are equal.
$iv$. Opposite angles are supplementary.
Of these statements:

A
$(ii)$ and $(iv)$ are correct
✓
$(ii)$ and $(iii)$ are correct
C
$(i)$ and $(iii)$ are correct
D
$(i)$ and $(iv)$ are correct

Answer

Correct option: B.

$(ii)$ and $(iii)$ are correct

When two straight lines intersect them, Adjacent angles are supplementary and opposite angles are equal.

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MCQ 1461 Mark

In the given figure, straight lines $AB$ and $CD$ interect at $O$.If $\angle\text{AOC}=\phi,\angle\text{BOC}=\theta$ and $\theta=3\phi,$ then $\phi=?$

A
$30^\circ$
B
$40^\circ$
✓
$45^\circ$
D
$60^\circ$

Answer

Correct option: C.

$45^\circ$

$\angle\text{AOD}=\angle\text{COB}=\theta$
$\angle\text{AOC}=\angle\text{BOD}=\phi$
Since the sum of the measures of the angles around a point is 360º,
$\angle\text{AOD}+\angle\text{COB}+\angle\text{AOC}+\angle\text{BOD}=360^\circ$
$\Rightarrow\theta+\theta+\phi+\phi=360$
$\Rightarrow2(\theta+\phi)=360$
$\Rightarrow\theta+\phi=180$
Given that $\theta=3\phi.$
So, $3\phi+\phi=180$
$\Rightarrow4\phi=180$
$\Rightarrow\phi=45$

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MCQ 1471 Mark

Which of the following statements is false?

A
Two straight lines can intersect only at one point.
B
A line segment can be produced to any desired length.
✓
Through a given point, only one straight line can be drawn.
D
Through two given points, it is possible to draw one and only one straight line.

Answer

Correct option: C.

Through a given point, only one straight line can be drawn.

This statement is false because we can draw infinitely many straight lines through a given point.

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MCQ 1481 Mark

In Fig. if $CP \| BQ$, then the measure of $x$ is:

A
$125^\circ$
✓
$130^\circ$
C
$175^\circ$
D
$105^\circ$

Answer

Correct option: B.

$130^\circ$

Given that,

$CP \| BQ$
Produce $CP$ to $E$
So, $PE \| BQ$ and $AB$ cuts them
$\angle\text{QBE}=\angle\text{QBA}=105^\circ$ (Corresponding angles)
In $\triangle\text{ECA}$
$\angle\text{CEA}+\angle\text{ECA}+\angle\text{EAC}=180^\circ$
$105^\circ+\angle\text{ECA}+25^\circ=180^\circ$
$\angle\text{ECA}=50^\circ$
$\angle\text{PCA}+\angle\text{ECA}=180^\circ$ (Linear pair)
$x + 50^\circ = 180^\circ$
$x = 130^\circ .$

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MCQ 1491 Mark

The number of triangles that can be drawn having angles as $50^\circ , 60^\circ $ and $70^\circ $ are:

A
Only one.
B
Two.
✓
Infinite.
D
None of these.

Answer

Correct option: C.

Infinite.

As we know similar triangles can be drawn for any given triangle.
These similar triangles will have the same angles as the original triangle (ie., $\angle50^\circ,\angle60^\circ$and $\angle70^\circ$) and will be infinite in number.

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MCQ 1501 Mark

In figure, if lines $l$ and $m$ are parallel, then the value of $x$, is:

✓
$35^\circ$
B
$55^\circ$
C
$65^\circ$
D
$75^\circ$

Answer

Correct option: A.

$35^\circ$

From figure,
$\angle\text{ACB}=180^\circ-\angle\text{ACD}=180^\circ-125^\circ=55^\circ$
Or
$\angle\text{BCA}=55^\circ$
In right $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{CAB}=180^\circ$
$\Rightarrow\ 90^\circ+55^\circ+\text{x}=180^\circ$
$\Rightarrow\ \text{x}=35^\circ$

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MCQ 1511 Mark

The measurement of Complete angle is:

A
$180^\circ$
B
$0^\circ$
✓
$360^\circ$
D
$90^\circ$

Answer

Correct option: C.

$360^\circ$

$x + y = 180^\circ $ (Linear Pair)
$w + z = 180^\circ $ (Linear Pair)
On adding above equations
We get, $x + y + w + z = 180^\circ + 180^\circ = 360^\circ $
Sum of all the angles around a point is $360^\circ .$

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MCQ 1521 Mark

In figure, $AB \| CD \| EF$ and $GH \| KL.$ The measure of $\angle\text{HKL},$ is:

A
$85^\circ$
B
$135^\circ$
✓
$145^\circ$
D
$215^\circ$

Answer

Correct option: C.

$145^\circ$

$GH \| KL$
$\Rightarrow\ \angle\text{GHK}=\angle\text{HKL}$ [Interior opposite angles]
Now,
$\angle\text{GHK}=\angle\text{GHD}+\angle\text{DHR}$
$=(180^\circ-\angle\text{GHC})+\angle\text{DHK}$
[$\angle\text{GHC}$ and $\angle\text{GHD}$ are supplementary]
$=180^\circ-60^\circ+25^\circ$
$\Rightarrow\ \angle\text{GHK}=145^\circ$
$\Rightarrow\ \angle\text{HKL}=\angle\text{GHK}=145^\circ$

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MCQ 1531 Mark

In fig if $x = 30^\circ $ then $y =$

A
$210^\circ$
B
$180^\circ$
C
$90^\circ$
✓
$150^\circ$

Answer

Correct option: D.

$150^\circ$

$x + y = 180^\circ $ (linear pair)
$x = 30^\circ$
$30^\circ + y = 180^\circ$
$y = 180^\circ - 30^\circ$
$y = 150^\circ .$

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MCQ 1541 Mark

In the adjoining figure, $\text{AB}\parallel\text{CD}$ and $\text{AB}\parallel\text{EF}$ The value of $x$ is:

A
$70^\circ$
B
$40^\circ$
✓
$60^\circ$
D
$50^\circ$

Answer

Correct option: C.

$60^\circ$

$\angle\text{FEC}+\angle\text{ECD}=180^\circ$ (sum of 2 supplimentary angles is 180°)
$\angle\text{FEC}+\angle180^\circ-150^\circ=30^\circ$
$\angle\text{X}=\angle\text{BCE}=\angle\text{ECD}$
$\angle\text{X}=30^\circ+30^\circ=60^\circ$

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MCQ 1551 Mark

Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is:

A
$45^\circ$
B
$30^\circ$
✓
$36^\circ$
D
None of these

Answer

Correct option: C.

$36^\circ$

Let one angle be $\theta$
Then, its complementary $=90-\theta$
According to question,
$2\theta=3(90-\theta)$
$=5\theta=270$
$\theta=54^\circ$
Then, $90-\theta^\circ=36^\circ$
Hence, the smaller angle is $36^\circ .$

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MCQ 1561 Mark

In the given figure, $\angle\text{OEB}=75^\circ,\angle\text{OBE}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ Then $\angle\text{ODC}=?$

A
$25^\circ$
✓
$30^\circ$
C
$35^\circ$
D
$20^\circ$

Answer

Correct option: B.

$30^\circ$

In $\triangle\text{OEB}$
$\angle\text{OEB}+\angle\text{EBO}+\angle\text{BOE}=180^\circ$ (Angle sum property)
$75^\circ+55^\circ+\angle\text{BOE}=180^\circ$
$\angle\text{BOE}=50^\circ$
$\angle\text{BOE}=\angle\text{COD}=50^\circ$ (Vertically opposite angle)
In $\triangle\text{ODC}$
$\angle\text{ODC}+\angle\text{DOC}+\angle\text{DCO}=180^\circ$
$\angle\text{ODC}=180^\circ-100^\circ-50^\circ$
$\angle\text{ODC}=30^\circ.$

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MCQ 1571 Mark

An exterior angle of a triangle is $105^\circ $ and its two interior opposite angles are equal. Each of these equal angles is:

A
$37\frac{1}{2}^\circ$
✓
$52\frac{1}{2}^\circ$
C
$75^\circ$
D
$72\frac{1}{2}^\circ$

Answer

Correct option: B.

$52\frac{1}{2}^\circ$

Let one of interior angle be $x^\circ $
Sum of two opposite interior angles = Exterior angle
$\therefore x^\circ + x^\circ = 105^\circ $ [ $\therefore$ Exterior angle $= 105^\circ$ (given)]
$\Rightarrow2\text{x}^\circ=105^\circ$
$\therefore\text{x}^\circ=\frac{105^\circ}{2}$
$\Rightarrow\text{x}^\circ=52\frac{1}{2}$
Hence, each angle of a triangle is $52\frac{1}{2}^\circ.$

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MCQ 1581 Mark

In the given figure $L1 \| L2$ and $\angle1=520.$ Find the measure of $\angle2:$

A
$38^\circ$
B
$48^\circ$
✓
$128^\circ$
D
$52^\circ$

Answer

Correct option: C.

$128^\circ$

$\angle1+\angle\text{a}=180^\circ$
$\angle1=52^\circ$
$52^\circ+\angle\text{a}=180^\circ$
$\angle\text{a}=180^\circ-52^\circ$
$\angle\text{a}=128^\circ$
$\angle\text{a}=\angle2$ (Corresponding angles)
Therefore $\angle2=128^\circ.$

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MCQ 1591 Mark

An exterior angle of a triangle is $110^\circ $ and its two interior opposite angles are equal. Each of these equal angles is:

A
$70^\circ$
B
$35^\circ$
✓
$55^\circ$
D
$27\frac{1}{2}^\circ$

Answer

Correct option: C.

$55^\circ$

Let the measure of each of the two equal interior opposite angles of the triangle be $x.$
In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.
$\therefore$ $x + x = 110^\circ$
$\Rightarrow 2x = 110^\circ$
$\Rightarrow x = 55^\circ$
Thus, the measure of each of these equal angles is $55^\circ .$

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MCQ 1601 Mark

In Fig., if lines $l$ and $m$ are parallel lines, then $x =$

A
$70^{\circ}$
✓
$40^{\circ}$
C
$100^{\circ}$
D
$30^{\circ}$

Answer

Correct option: B.

$40^{\circ}$

Given that,
$l \| m$
Let, $l \| m$ and transversal cuts them and
$\angle1=70^\circ$
$\angle3=20^\circ$
$\angle4=30^\circ$
$\angle1+\angle2=180^\circ$ (Interior angle)
$\angle2=110^\circ\text{(i})$
$\angle2=\angle5$ (Vertically opposite angle)
$\angle5=110^\circ\text{(ii)}$
$\angle5+\angle3+\angle4=180^\circ$(Sum of angles of a triangle is $180^{\circ}$)
$110^{\circ} + x + 30^{\circ} = 180^{\circ}$
$x = 40^{\circ}$.

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MCQ 1611 Mark

In the given figure, $AB \| CD$. If $\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ$ then $\angle\text{OCD}=?$

A
$150^\circ$
B
$80^\circ$
✓
$130^\circ$
D
$100^\circ$

Answer

Correct option: C.

$130^\circ$

Draw $OE \| AB \| CD$
Now, $OE \| AB$ and $OA$ is the transversal.
$\therefore\angle\text{OAB}+\angle\text{AOE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow\angle\text{OAB}+\angle\text{AOC}+\angle\text{COE}=180^\circ$
$\Rightarrow100^\circ+30^\circ+\angle\text{COE}=180^\circ$
$\Rightarrow\angle\text{COE}=50^\circ$
Also,
$OE \| CD$ and $OC$ is the transversal.
$\therefore\angle\text{OCD}+\angle\text{COE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
$\Rightarrow\angle\text{OCD}=130^\circ.$

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MCQ 1621 Mark

Write the correct answer in the following: If one of the angles of a triangle is $130^\circ $, then the angle between the bisectors of the other two angles can be.

A
$50^\circ$
B
$65^\circ$
C
$145^\circ$
✓
$155^\circ$

Answer

Correct option: D.

$155^\circ$

In $\Delta\text{ABC},$ we have $\angle\text{A}=130^\circ$OB and $OC$ are the bisectors of the angles $B$ and $C.$
Let $\angle\text{OBC}=\angle\text{OBA}=\text{x}\ \text{and}\angle\text{OCB}=\angle\text{OCA}=\text{y}$
In $\Delta\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow130^\circ+2\text{x}+2\text{y}=180^\circ$
$\Rightarrow\text{x}+\text{y}=25^\circ$
$\text{i.e}\angle\text{OBC}+\angle\text{OCB}=25^\circ$
Now, In $\Delta\text{BOC}$
$\angle\text{BOC}=180^\circ-\big(\angle\text{OBC}+\angle\text{OCB}\big)$ (Angle sum Property)
$=180^\circ-25^\circ=155^\circ$
Hence, $(d)$ is the correct answer.

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MCQ 1631 Mark

Consider the following statement:When two straight lines intersect:
$i$. Adjacent angles are complementary
$ii$. Adjacent angles are supplementary
$iii$. Opposite angles are equal
$iv$. Opposite angles are supplementary
Of these statements

A
$(i)$ and $(ii)$ are correct
✓
$(ii)$ and $(iii)$ are correct
C
$(i)$ and $(iv)$ are correct
D
$(ii)$ and $(iv)$ are correct

Answer

Correct option: B.

$(ii)$ and $(iii)$ are correct

Let two lines $AB$ and $CD$ intersect each other at $O.$
Now,
We can see from fogure any two adjacent angles
$\angle\text{AOD}$ and $\angle\text{DOB},\angle\text{DOB}$ and $\angle\text{BOC}$ etc are supplementary because their sum is $180^\circ .$
$\angle\text{AOD}+\angle\text{DOB}=180^\circ$
$\angle\text{DOB}+\angle\text{BOC}=180^\circ$
So two adjacent angles are always supplementary.
Now,
Two opposite angle like $\angle\text{AOC}$ and $\angle\text{DOB},\angle\text{AOD}$ and $\angle\text{COB}$ are always equal to each other as they are vertically opposite angles
$\angle\text{AOC}=\angle\text{DOB}$
$\angle\text{AOD}=\angle\text{COB}$
Hence statement $(ii)$ and $(iii)$ are correct.

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MCQ 1641 Mark

In figure, lines $I_1 \| I_2$. The value of $x$ is:

A
$40^{\circ}$
✓
$30^{\circ}$
C
$50^{\circ}$
D
$70^{\circ}$

Answer

Correct option: B.

$30^{\circ}$

40$^{\circ}$ + x = 70$^{\circ}$ (exterior angle)
$\angle\text{x}=70^\circ-40^\circ$
$\angle\text{x}=30^\circ.$

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MCQ 1651 Mark

Write the correct answer in the following: If one angle of a triangle is equal to the sum of the other two angles, then the triangle is.

A
An isosceles triangle.
B
An obtuse triangle.
C
An equilateral triangle.
✓
An equilateral triangle.

Answer

Correct option: D.

An equilateral triangle.

Let the angles of $\Delta\text{ABC}\ \text{be}\ \angle\text{A},\angle\text{B}\ \text{and}\ \angle\text{C}$
Given that $\angle\text{A}=\angle\text{B}+\angle\text{C}.....(1)$
But, in any $\Delta\text{ABC},\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ....(2)$[Angles sum property of triangle]
From equation (1) and (2), we get
$\angle\text{A}+\angle\text{A}=180^\circ\Rightarrow2\angle\text{A}=180^\circ\Rightarrow\angle\text{A}\frac{180^\circ}{2}=90^\circ$
$\Rightarrow\angle\text{A}=90^\circ$
Hence, the triangle is a right triangle and option $(d)$ is correct.

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MCQ 1661 Mark

In the adjoining figure, $BE$ and $CE$ are bisectors of $\angle\text{ABC}$ and $\angle\text{ACD}$ respectively. If $\angle\text{BEC}=25^\circ$ then $\angle\text{BAC}$ is equal to :

✓
$50^\circ$
B
$25\frac{1}{2}^0$
C
$65^\circ$
D
$12\frac{1}{2}^0$

Answer

Correct option: A.

$50^\circ$

In $\triangle\text{BEC}$
$\angle\text{BEC}+\angle\text{EBC}+\angle\text{ECD}$ (Exterior angle property)
$\angle\text{BEC}=\angle\text{ECD}-\angle\text{EBC}$
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}=\angle\text{ACD}$
$\angle\text{ABC}+2\angle\text{EBC}=2\angle\text{ECD}$
$\angle\text{ABC}=2(\angle\text{ECD}-\angle\text{EBC})$
$\angle\text{ABC}=2(\angle(\text{BEC}))$
$\angle\text{ABC}=50^\circ$

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MCQ 1671 Mark

Two complementary angles are such that twich the measure of the one is equal to three times the measure of the other. The larger of the two measure.

A
$72^\circ$
✓
$54^\circ$
C
$63^\circ$
D
$36^\circ$

Answer

Correct option: B.

$54^\circ$

Let the measure of each angle be $x^\circ $ and $(90 - x)^\circ .$
According to the given condition,
$2x = 3(90 - x)$
$\Rightarrow 2x = 270 - 3x$
$\Rightarrow 5x = 270$
$\Rightarrow x = 54^\circ$
So,$ (90 - x)^\circ = (90 - 54)^\circ = 36^\circ$
So, the larger of the two angles is $54^\circ .$

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MCQ 1681 Mark

In figure, if $I_1 \| I_2$, what is $x + y$ in terms of w and $z?$

✓
$180 - w + z$
B
$180 + w - z$
C
$180 - w - z$
D
$180 + w + z$

Answer

Correct option: A.

$180 - w + z$

Let angle supplement of $w^{\circ}$ be $a^{\circ}$.
$\Rightarrow \mathrm{a}^{\circ}=180^{\circ}-\mathrm{w}^{\circ}$
Now,
$a^{\circ}=x^{\circ} \text { [Alternate opposite angles] }$
$\Rightarrow x^{\circ}=180-w^{\circ} \ldots(1)$
Now,
$y^{\circ}=z^{\circ} \ldots \text { (2) [Alternate angles] }$
Adding $(1)$ and $(2)$, we get
$x^{\circ}+y^{\circ}=180^{\circ}-w^{\circ}+z^{\circ}$

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MCQ 1691 Mark

In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=(3\text{x}-10)^\circ,\angle\text{COD}=50^\circ$ and $\angle\text{BOD}=(\text{x}+20)^\circ$ then $\angle\text{AOC}=?$

A
$40^\circ$
B
$60^\circ$
✓
$80^\circ$
D
$50^\circ$

Answer

Correct option: C.

$80^\circ$

Since $AOB$ is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow(3\text{x}-10)+50+(\text{x}+20)=180^\circ$
$\Rightarrow4\text{x}+60=180$
$\Rightarrow4\text{x}=120$
$\Rightarrow\text{x}=30$
So, $\angle\text{AOC}=3\text{x}-10=3(30)-10=80^\circ$

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MCQ 1701 Mark

The sides of a quadrilateral are extended in order to form 4 exterior angles. The sum of these exterior angles is:

✓
$360^\circ$
B
$180^\circ$
C
$720^\circ$
D
$90^\circ$

Answer

Correct option: A.

$360^\circ$

As seen from the above figure:
$\angle1+\angle2=180^\circ$ (Linear Pair)
Similarly $\angle3+\angle4=180=\angle5+\angle6=\angle7+\angle8=180^\circ$
Adding up all of the above
$\angle1+\angle2+\angle3+\angle4+\angle5+\angle6+\angle7+\angle8=180\times4=720^\circ $
But we know that the angles of a quadrilateral add up to $360^\circ$
Therefore $\angle2+\angle3+\angle5+\angle7=360^\circ$
Substituting in the above equation we get:
$\angle1+\angle4+\angle6+\angle8+360^\circ=720^\circ$
Therefore $\angle1+\angle4+\angle6+\angle8=360^\circ$
Hence the sum of all exterior angles $= 360^\circ .$

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MCQ 1711 Mark

In the adjoining figure, $AOB$ is a straight line. If $x : y : z = 4 : 5 : 6$, then $y =?$

A
$72^{\circ}$
✓
$60^{\circ}$
C
$48^{\circ}$
D
$80^{\circ}$

Answer

Correct option: B.

$60^{\circ}$

Let $\angle\text{AOC}=\text{x}^\circ=(4\text{a})^\circ,\angle\text{COD}=\text{y}^\circ=(5\text{a})^\circ$ and $\angle\text{BOD}=\text{z}^\circ=(6\text{a})^\circ$
Then, we have
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$ [Since $AOB$ is a straight line]
$\Rightarrow 4a + 5a + 6a= 180^{\circ}$
$\Rightarrow 15a = 180^{\circ}$
$\Rightarrow a = 12^{\circ}$
$\therefore y = 5 \times a = 5 \times 12^\circ = 60^{\circ}$.

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MCQ 1721 Mark

In the figure, $AB \| CD$. If $\angle\text{APQ}=70 ^\circ$ and $\angle\text{PRD}=120^\circ$ then $\angle\text{QPR}=?$

✓
$55^\circ$
B
$60^\circ$
C
$40^\circ$
D
$35^\circ$

Answer

Correct option: A.

$55^\circ$

Since $AB \| CD,$
$\angle\text{APQ}=\angle\text{PQR}=70^\circ$ (Alternate angles)
$\Rightarrow\angle\text{PRD}=\angle\text{PQR}+\angle\text{QPR}$ (Exterior angle is equal to sum of the remote interior angles)
$\Rightarrow120^\circ=70^\circ+\angle\text{QPR}$
$\Rightarrow\angle\text{QPR}=506^\circ$

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MCQ 1731 Mark

In figure, if $AB \| CD$, then $x =$

✓
$100$
B
$105$
C
$110$
D
$114$

Answer

Correct option: A.

$100$

Extending line $BA$ and $CP$ to meet at point $E.$
$\angle\text{APE}=180^\circ-148^\circ=32^\circ$
$\angle\text{EAP}=180^\circ-132^\circ=48^\circ$
$\angle\text{AEP}=\text{x}^\circ$ [(Correspondence angles) because $AB \| CD$ cut by transverse $EC]$
Now, in $\triangle\text{APE}$
$\angle\text{APE}+\angle\text{EAP}+\angle\text{AEP}=180^\circ$
$\Rightarrow\ 32^\circ+48^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow\ \text{x}^\circ=100$

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MCQ 1741 Mark

One angle is equal to three times its supplement. The measure of the angle is:

A
$130^\circ$
✓
$135^\circ$
C
$90^\circ$
D
$120^\circ$

Answer

Correct option: B.

$135^\circ$

Let the required angle be ${\theta}.$
Then, measure of its supplement $180^\circ-\theta$
According to question, we have
$\theta=3(180-\theta)$
$\Rightarrow\ \theta=540^\circ-3\theta$
$\Rightarrow\ 4\theta=540^\circ $
$\Rightarrow\ \theta=135^\circ$

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MCQ 1751 Mark

In the adjoining figure, the bisectors of $\angle\text{CBD}$ and $\angle\text{BCE}$ meet at the point O. If $\angle\text{BAC}=70^\circ$ then $\angle\text{BOC}$ is equal to:

A
$11^\circ$
✓
$55^\circ$
C
$70^\circ$
D
$35^\circ$

Answer

Correct option: B.

$55^\circ$

$\angle\text{BOC}=90^\circ-\frac{1}{2 }\angle\text{BAC}$
$\angle\text{BOC}=90^\circ-35^\circ=55^\circ$

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MCQ 1761 Mark

Two planes intersect each other to form a:

A
Plane.
B
Point.
C
Angle.
✓
Straight line.

Answer

Correct option: D.

Straight line.

As can be seen from the above diagram, the two planes $"P"$ and $"Q"$ are intersecting in a line, which is $AB.$

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MCQ 1771 Mark

In the given figure, $AB \I CD$. If $\angle\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ.$ then $\angle\text{OCD}=?$

✓
$80^\circ$
B
$100^\circ$
C
$130^\circ$
D
$150^\circ$

Answer

Correct option: A.

$80^\circ$

Extend line $CD$ which intersect $AO$ at $M.$
(Corresponding angle)
$\angle\text{MOC}+\angle\text{CMO}=\angle\text{DCO}$ (exterior angle is equal to the sum of two opposite interior angles)
$\angle\text{DCO}=100^\circ+30^\circ=130^\circ$

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MCQ 1781 Mark

In the given figure, $AB$ is a mirror, $PQ$ is the incident ray and $QR$ is the reflected ray. If $\angle\text{PQR}=108^\circ$ then $\angle\text{AQP}=?$

A
$72^\circ$
B
$18^\circ$
✓
$36^\circ$
D
$54^\circ$

Answer

Correct option: C.

$36^\circ$

We know that, angle of incidance = angle reflection.
that is, $\angle\text{AQP}=\angle\text{BQR}$
Since $AOB$ is a straight line,
$\angle\text{AQP}+\angle\text{BQR}+\angle\text{PQR}=180^\circ$
$\Rightarrow\angle\text{AQP}+\angle\text{AQP}+\angle\text{PQR}=180^\circ$
$\Rightarrow2\angle\text{AQP}=72$
$\Rightarrow \angle\text{AQP}=36^\circ$

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MCQ 1791 Mark

Measurement of reflex angle is:

A
Between $0^\circ $ and $90^\circ $
B
$90^\circ$
✓
Between $180^\circ $ and $360^\circ $
D
Between $90^\circ $ and $180^\circ $

Answer

Correct option: C.

Between $180^\circ $ and $360^\circ $

Let $x$ be the angle
then its reflex angle is $360^\circ - x$
and in any triangle the angle lies between $0$ to $180^\circ $

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MCQ 1801 Mark

One angle is equal to three times its supplement. The measure of the angle is:

A
$130^\circ$
B
$120^\circ$
✓
$135^\circ$
D
$90^\circ$

Answer

Correct option: C.

$135^\circ$

Let the required angle be $x$
Supplement $= 180^\circ - x$
According to question,
$x = 3 (180^\circ - x)$
$x = 540^\circ - 3x$
$x = 135^\circ .$

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MCQ 1811 Mark

In the given figure, $\angle\text{BAC}=40^\circ,\angle\text{ACB}=90^\circ$ and $\angle\text{BED}=100^\circ$ Then $\angle\text{BDE}=?$

A
$25^\circ$
B
$40^\circ$
C
$50^\circ$
✓
$30^\circ$

Answer

Correct option: D.

$30^\circ$

In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$(Angle sum property)
$\angle\text{ABC}=180^\circ-90^\circ-40^\circ$
$\angle\text{ABC}=50^\circ$
In $\triangle\text{BED}$
$\angle\text{BED}+\angle\text{EBD}+\angle\text{BDE}=180^\circ$(Angle sum property)
$\angle\text{EBD}=180^\circ-50^\circ-100^\circ$
$\angle\text{EBD}=30^\circ$

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MCQ 1821 Mark

In the adjoining figure, $\angle\text{a}$ and $\angle\text{g}$ are called:

A
Co-interior angles.
B
Corresponding angles.
✓
Alternate exterior angles.
D
Alternate interior angles.

Answer

Correct option: C.

Alternate exterior angles.

$\angle\text{a}$ and $\angle\text{g}$ are on alternate side and are exterior.

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MCQ 1831 Mark

In figure, if $AB \| CF, CD \| EF$, then the value of $x$ is:

A
$140^\circ$
B
$120^\circ$
C
$100^\circ$
✓
$110^\circ$

Answer

Correct option: D.

$110^\circ$

In the above figure, $\angle\text{B}=\angle\text{BCF}$ (Alternate Interior angles)
Now $\angle\text{FCA}=\angle\text{BCA}+\angle\text{FCB}$
$= 60^\circ + 50^\circ = 110^\circ$
Now $\angle\text{FCA}=\angle\text{x} $ (Alternate interior angles)
Therefore $\angle\text{x}=110^\circ.$

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MCQ 1841 Mark

The sides $BC, CA$ and $AB$ of $\triangle\text{ABC}$ have been produced to $D, E$ and $F$ respectively as shown in the figure, forming exterior angles $\angle\text{ACD},\angle\text{BAE}$ and $\angle\text{CBF}.$ Then, $\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=?$

A
$240^\circ$
✓
$360^\circ$
C
$300^\circ$
D
$320^\circ$

Answer

Correct option: B.

$360^\circ$

We know that the exteior angle is sum of opposite interior angle
$\angle\text{ACD}=\angle1+\angle2\text{ (i)}$
$\angle\text{BAE}=\angle3+\angle2\text{ (ii)}$
$\angle\text{CBF}=\angle1+\angle3\text{ (iii)}$
$ADD$ Equation $(i), (ii)$ and $(iii)$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times(\angle1+\angle2+\angle3)$
$\angle1+\angle2+\angle3=180^\circ\text{(ASP)}$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times180^\circ$
$\angle\text{ACD} +$ We know that the exteior angle is sum of opposite interior angle
$\angle\text{ACD}=\angle1+\angle2\text{ (i)}$
$\angle\text{BAE}=\angle3+\angle2\text{ (ii)}$
$\angle\text{CBF}=\angle1+\angle3\text{ (iii)}$
$ADD$ Equation $(1i), (ii)$ and $(iii)$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times(\angle1+\angle2+\angle3)$
$\angle1+\angle2+\angle3=180^\circ\text{(ASP)}$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times180^\circ$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=360^\circ.$

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MCQ 1851 Mark

In the adjoining figure, $AB \| CD$ and $AB \| EF$. If $\text{EA}\bot\text{BA}$ and $\angle\text{BEF}=55^\circ,$ then the values of $x, y$ and $z:$

A
$35^\circ , 125^\circ , 120^\circ$
✓
$125^\circ , 125^\circ , 35^\circ$
C
$60^\circ , 60^\circ , 60^\circ$
D
$120^\circ , 130^\circ , 25^\circ$

Answer

Correct option: B.

$125^\circ , 125^\circ , 35^\circ$

$x + 55 = 180^\circ $ (Sum of supplementary angles or co-interior angles)$x = 125^\circ$
$x = y = 125^\circ $ (Corresponding angles)
$\text{z}+\angle\text{EAB}=\text{y} $ (Exterior angle property)
$z = 125^\circ - 90^\circ = 35^\circ .$

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MCQ 1861 Mark

The number of line segments determined by three given non$-$collinear points is:

A
Four.
B
Infinitely many.
C
Two.
✓
Three.

Answer

Correct option: D.

Three.

Three because non$-$collinear points means the point does not lies in a same line.

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MCQ 1871 Mark

In the figure, $POQ$ is a line. The value of $x$ is:

A
$30^\circ$
✓
$20^\circ$
C
$35^\circ$
D
$25^\circ$

Answer

Correct option: B.

$20^\circ$

Since, $POQ$ is a line segment.
$\therefore\angle\text{POQ}=180^\circ$
$\Rightarrow\angle\text{POA}+\angle\text{AOB}+\angle\text{BOQ}=180^\circ$
$\Rightarrow40^\circ+4\text{x}+3\text{x}=180^\circ$
[Putting $\angle\text{POA}=40^\circ,\angle\text{AOB}=4\text{x}$ and $\angle\text{BOQ}=3\text{x}$]
$\Rightarrow7\text{x}=180^\circ-40^\circ$
$\Rightarrow7\text{x}=140^\circ$
$\text{x}=\frac{140^\circ}{7}$
$\text{x}=20^\circ.$

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MCQ 1881 Mark

In Fig. the value of $y$ is:

A
$20^{\circ}$
B
$60^{\circ}$
✓
$30^{\circ}$
D
$45^{\circ}$

Answer

Correct option: C.

$30^{\circ}$

$3x + y + 2x = 180^{\circ}$(Linear pair)
$5x + y = 180^{\circ} (i)$
From figure,
$y = x$ (Vertically opposite angles)
Using it in $(i)$, we get
$5x + x = 180^{\circ}$
$6x = 180^{\circ}$
$x = 30^{\circ}$
Thus,
$y = x = 30^{\circ}$.

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MCQ 1891 Mark

An angle which measures more than $180^{\circ}$ but less than $360^{\circ}$, is called.

A
An acute angle.
B
An obuse angle.
C
A straight angle.
✓
A reflex angle.

Answer

Correct option: D.

A reflex angle.

An angle which measures more than $180^{\circ}$ but less than $360^{\circ}$is called a reflex angle.

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MCQ 1901 Mark

In Fig., if $l \| m,$ then $x =$

A
$40^{\circ}$
B
$25^{\circ}$
C
$65^{\circ}$
✓
$105^{\circ}$

Answer

Correct option: D.

$105^{\circ}$

Given that,
$l \| m$ and n cuts them
Let,
$\angle1=65^\circ$
$\angle2=\text{x}$
$\angle3=40^\circ$
$\angle1+\angle4=65^\circ$ (Alternate angle) $(i)$
$\angle3+\angle4+\angle5=180^\circ$(Angle sum property)
$40^\circ+165^\circ+\angle5=180^\circ$
$\angle5=75^\circ$
Now,
$\angle2+\angle5=180^\circ$ (Linear pair)
$x + 75^{\circ}$ = 180$^{\circ}$
$x = 105^{\circ}$.

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MCQ 1911 Mark

Angles of a triangle are in the ratio $2 : 4 : 3$. The smallest angle of the triangle is:

A
$20^\circ$
✓
$40^\circ$
C
$60^\circ$
D
$80^\circ$

Answer

Correct option: B.

$40^\circ$

Given, the ratio of angles of a triangle is $2 : 4 : 3.$
Let the angles of a triangle be $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$
$\angle\text{A}=2\text{x}, \angle\text{B}=4\text{x}$
$\angle\text{C}=3\text{x},\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
[sum of all the angles of a triangle is 180º]
$2\text{x}+4\text{x}+3\text{x}=180^\circ$
$9\text{x}=180^\circ$
$\text{x}=\frac{180^\circ}{9}=20^\circ$
$\angle\text{A}=2\text{x}=2\times20^\circ=40^\circ$
$\angle\text{B}=4\text{x}=2\times20^\circ=80^\circ$
$\angle\text{C}=3\text{x}=3\times20^\circ=60^\circ$
Hence, the smallest angle of a triangle is $40^\circ .$

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MCQ 1921 Mark

The number of lines that can pass through a given point is:

A
Only one
B
Two
C
One
✓
Infinity

Answer

Correct option: D.

Infinity

As seen from the above image, any number of lines can be drawn through a given point.
Hence, the answer may be given as $"$Infinity$"$.

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MCQ 1931 Mark

If two angles are complements of each other then each angle is:

A
A straight angle.
B
A reflex angle.
✓
An acute angle.
D
An obtuse angle.

Answer

Correct option: C.

An acute angle.

If two angles are complements of each other, that is, the sum of their measures is $90^\circ $, then each angle is an acute angle.

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MCQ 1941 Mark

Write the correct answer in the following: In Fig. if $\text{OP}||\text{RS},$ $\angle\text{OPQ}=110^\circ$ and $\angle\text{QRS}=130^\circ,$ then $\angle\text{PQR}$ is equal to,

A
$40^\circ$
B
$50^\circ$
✓
$60^\circ$
D
$70^\circ$

Answer

Correct option: C.

$60^\circ$

In the given figure, producing $OP$, to interscet $RQ$ at $X.$
Since, $\text{OP}||\text{RS}$ and RX is a transversal.
So, $\angle\text{RXP}=\angle\text{XRS}$

$\Rightarrow\angle\text{RXP}=130^\circ$$\big[\because\angle\text{QRS}=130^\circ(\text{given})\big]....(\text{i})$
Now, $RQ$ is a line segment.
So,$\angle\text{PXQ}+\angle\text{RXP}=180^\circ$ [linear pair axiom]
$\Rightarrow\angle\text{PXQ}=180^\circ-\angle\text{RXP}=180^\circ-130^\circ$ [from eq. $(i)]$
$\Rightarrow\angle\text{PXQ}=50^\circ$
In $\Delta\text{PQX},$ $\angle\text{OPQ}$ is an exterior angle,
$\therefore\angle\text{OPQ}=\angle\text{PXQ}+\angle\text{PQX}$
[$\because\ $exterior asngle = sum of two opposite interior angles]
$110^\circ=50^\circ+\angle\text{PQX}$
$\angle\text{PQX}=110^\circ-50^\circ$
$\angle\text{PQR}=60^\circ$ $[\because\ \angle\text{PQX}=\angle\text{PQR}]$

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MCQ 1951 Mark

In figure, if $AB \| HF$ and $DE \| FG$, then the measure of $\angle\text{FDE}$ is:

A
$108^\circ$
✓
$80^\circ$
C
$100^\circ$
D
$90^\circ$

Answer

Correct option: B.

$80^\circ$

AB || HF and $\angle\text{CFH}=28^\circ$ [Given]
$\angle\text{CFH}=\angle\text{FDA}$ [Correspondence angels are equal]
$\angle\text{FDA}=28^\circ$
Now,
$\angle\text{FDA}+\angle\text{FDE}+\angle\text{EDB}=180^\circ$
$\Rightarrow\ 28^\circ+\angle\text{FDE}+72^\circ=180^\circ$
$\Rightarrow\ \angle\text{FDE}=80^\circ$

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MCQ 1961 Mark

In the given figure, straight lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{BOD}=130^\circ$ then $\angle\text{AOD}=?$

✓
$115^\circ$
B
$125^\circ$
C
$110^\circ$
D
$65^\circ$

Answer

Correct option: A.

$115^\circ$

We have,
$\angle\text{AOC}=\angle\text{BOD}$ [Vertically-Opposite Angles]
$\therefore\angle\text{AOC}+\angle\text{BOD}=130^\circ$
$\Rightarrow\angle\text{AOC}+\angle\text{AOC}=130^\circ [\therefore\angle\text{AOC}=\angle\text{BOD}]$
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Now,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$ [$\because$ COD is a straight line]
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOC}=115^\circ.$

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MCQ 1971 Mark

In the given figure, straight lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}=\phi,\angle\text{BOC}=\theta$ and $\theta=3\phi,$ then $\phi=?$

✓
$45^\circ$
B
$30^\circ$
C
$60^\circ$
D
$40^\circ$

Answer

Correct option: A.

$45^\circ$

We have.
$\theta+\phi=180^\circ$ [$\because AOD$ is a straight line]
$\Rightarrow3\phi+\phi=180^\circ [\because\theta=3\phi]$
$\Rightarrow4\theta=180^\circ$
$\Rightarrow\phi=45^\circ.$

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MCQ 1981 Mark

If two lines intersect each other then:

A
Corresponding angles are equal.
B
Alternate interior angles are equal.
C
Co-interior angles are equal.
✓
Vertically opposite angles are equal.

Answer

Correct option: D.

Vertically opposite angles are equal.

$\angle\text{A}+\angle\text{B}=180^\circ ($Linear Pair$)$
$\angle\text{B}+\angle\text{C}=180 ($Linear Pair$)$
On equating above equations, we get
$\angle\text{A}+\angle\text{B}=\angle\text{B}+\angle\text{C}$
$\angle\text{A}=\angle\text{C}$
Similarly, $\angle\text{B}=\angle\text{D}$

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MCQ 1991 Mark

In the given figure, $AB \| CD \| EF,$ $\text{EA }\bot\text{ AB}$ and BDE is the transversal such that $\angle\text{DEF}=55^\circ,$ Then $\angle\text{AEB}=?$

✓
$35^\circ$
B
$55^\circ$
C
$45^\circ$
D
$25^\circ$

Answer

Correct option: A.

$35^\circ$

$\text{EA }\bot\text{ AB}$
$\angle\text{AEF}=90^\circ$
$\angle\text{AEF}=\angle\text{BEF}+\angle\text{AEB}$
$\angle\text{BEF}+\angle\text{AEB}=90^\circ$
$\angle\text{BEF}=55^\circ$
$55^\circ+\angle\text{AEB}=90^\circ$
$\text{AEB}=90^\circ-55^\circ$
$\angle\text{AEB}=35^\circ.$

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MCQ 2001 Mark

In the given figure, $BO$ and $CO$ are the bisectors of $\angle\text{B}$ and $\angle\text{C}$ respectively. If $\text{A}=50^\circ,$ then $\angle\text{BOC}=?$

A
$130^\circ$
B
$120^\circ$
C
$100^\circ$
✓
$115^\circ$

Answer

Correct option: D.

$115^\circ$

In $\triangle\text{ABC}$
$\text{2x}+\text{2y}+\angle\text{A}=180^\circ$ (Angle sum property)
$\text{x}+\text{y}+(\frac{\angle\text{A}}{2})=90^\circ$
$\text{x}+\text{y}=90^\circ-(\frac{\text{A}}{2})$(1)
In $\triangle\text{BOC},$ we have
$\text{x}+\text{y}+\angle\text{BOC}=180^\circ$
$90^\circ- (\frac{\angle\text{A}}{2})+\angle\text{BOC}=180^\circ$ [From (1)]
$\text{BOC}=180^\circ-90^\circ+(\frac{\text{A}}{2})$
$\angle\text{BOC}=90^\circ+(\frac{\text{A}}{2})$
$\angle\text{BOC}=90^\circ+25^\circ=115^\circ.$

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MCQ 2011 Mark

In Fig. $AOB$ is a straight line. If $\angle\text{AOC}+\angle\text{BOD}=85^\circ,$ then $\angle\text{COD}=$

A
$90^\circ$
B
$85^\circ$
✓
$95^\circ$
D
$100^\circ$

Answer

Correct option: C.

$95^\circ$

Given,
$AOB =$ Straight line
$\angle\text{AOC}+\angle\text{BOD}=85^\circ$
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ $(Linear pair)
$85^\circ+\angle\text{COD}=180^\circ$
$\angle\text{COD}=95^\circ.$

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MCQ 2021 Mark

If one of the angles of a triangle is $130^{\circ}$, then the angle between the bisectors of the other two angles can be:

A
$145^{\circ}$
B
$155^{\circ}$
✓
$50^{\circ}$
D
$65^{\circ}$

Answer

Correct option: C.

$50^{\circ}$

Let angles of a triangle be $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$

In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ $[sum of all interior angles of a triangle is $180^{\circ}$]
$\Rightarrow12\angle\text{A}+12\angle\text{B}+12\angle\text{C}=180^\circ2=90^\circ$[dividing both sides by $2]$
$\Rightarrow12\angle\text{B}+12\angle\text{C}=90^\circ−12\angle\text{A} $
$[\because\text{In}\triangle\text{OBC},\angle\text{OBC}+\angle\text{BCO}+\angle\text{COB}=180^\circ]$
⇒ Since, $\angle\text{B}2+\angle\text{C}2+\angle\text{BOC}=180^\circ$ as $BO$ and $OC$ are the angle bisectors of $\angle\text{ABC} $ and $\angle\text{BCA},$ respectively
$\Rightarrow180^\circ−\angle\text{BOC}=90^\circ−12\angle\text{A}$
$\therefore\angle\text{BOC}=180^\circ-90^\circ+12\angle\text{A}$
= $90^{\circ}+12 \times 130^{\circ}=90^{\circ}+65^{\circ}[\therefore\angle\text{A}=130^\circ(\text{given)}]$
$= 155^{\circ}$
Hence, the required angle is 155$^{\circ}$.

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MCQ 2031 Mark

If two angles are supplementary and the larger is $20^{\circ}$ less then three times the smaller, then the angles are:

A
$72\frac{1}{2}^0,17\frac{1}{2}^0$
B
$140^0,40^0$
✓
$130^0,50^0$
D
$62\frac{1}{2}^0,27\frac{1}{2}^0$

Answer

Correct option: C.

$130^0,50^0$

Let the two supplimentary angles be $\mathrm{x}^0$ and $180^{\circ}-\mathrm{x}^0$
Let $180^{\circ}-x$ be the larger angle
$180^{\circ}-x=3 x-20^0$
$4 x=200^{\circ}$
$x=50^{\circ}$
So the angles are $50^{\circ}$ and $130^{\circ}$

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MCQ 2041 Mark

In $\triangle\text{ABC, }\text{BD}\perp\text{AC, }\angle\text{CAE} = 30^\circ$ and $\angle\text{CBD}=40^\circ.$ Then $\angle\text{AEB}=?$

A
$70^\circ$
B
$50^\circ$
C
$60^\circ$
✓
$80^\circ$

Answer

Correct option: D.

$80^\circ$

In $BDC$
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{DBC}=180^\circ$
$\text{BD}\perp\text{AC}$
$\angle\text{BCD}=90^\circ,\angle\text{DBC}=40^\circ$
$90^\circ+\angle\text{BCD}+40^\circ=180^\circ$
$\angle\text{BCD}+130^\circ=180^\circ$
$\angle\text{BCD}=180^\circ-130^\circ$
$\angle\text{BCD}=50^\circ$
$\angle\text{AEB}=\angle\text{CAE}+\angle\text{C}$ (exterior angle)
$\angle\text{CAE}=30^\circ$
$\angle\text{C}=50^\circ$
$\angle\text{AEB}=30^\circ+50^\circ$
$\angle\text{AEB}=80^\circ.$

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MCQ 2051 Mark

Two complementary angles are such that twice the measure of the one is equal to three times the measure of the other. The larger of the two measures.

A
$72^\circ$
B
$36^\circ$
✓
$54^\circ$
D
$63^\circ$

Answer

Correct option: C.

$54^\circ$

Let the measure of the required angle be $x^\circ$
Then, the measure of its complement will be $(90 - x)^\circ$
Therefore, $2x = 3 (90 - x)$
$\Rightarrow 2x = 270 - 3x$
$\Rightarrow 5x = 270$
$\Rightarrow x = 54^\circ .$

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MCQ 2061 Mark

In figure, if $l \| m$, then $x =$

✓
$105^\circ$
B
$65^\circ$
C
$40^\circ$
D
$25^\circ$

Answer

Correct option: A.

$105^\circ$

From figure,
$\angle\text{AGE}=\angle\text{FGB}$ [Opposite angles]
$\Rightarrow\ \angle\text{FGB}=65^\circ$
Also,
$\angle\text{FGB}=\angle\text{HJI}$ [Corresponding angle]
$\Rightarrow\ \angle\text{HJI}=65^\circ$
Now, in $\angle\text{HJI},$
$\angle\text{HJI}+\angle\text{JIH}+\angle\text{IHJ}=180^\circ$
$\Rightarrow\ 65^\circ+40^\circ+\angle\text{IHJ}=180^\circ$
$\Rightarrow\ \angle\text{IHJ}=180^\circ-65^\circ-40^\circ=75^\circ$
Now,
$\text{x}=180^\circ-\angle\text{IHJ}=180^\circ-75^\circ$
$=105^\circ$

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MCQ 2071 Mark

In Fig. if $\frac{\text{y}}{\text{x}}=5$ and $\frac{\text{z}}{\text{x}}=4,$ then the value of $x$ is:

A
$15^\circ$
✓
$18^\circ$
C
$12^\circ$
D
$8^\circ$

Answer

Correct option: B.

$18^\circ$

In the given figure, we have $x^\circ , y^\circ $ and $z^\circ $ forming a linear pair, therefore these must be supplementary.
That is,
$x + y + z = 180^\circ (1)$
Also,
$\frac{\text{y}}{\text{x}}=5$
$y = 5x (2)$
And
$\frac{\text{z}}{\text{x}}=4$
$z = 4x (3)$
Substituting $(ii)$ and $(iii)$ in $(i),$ we get:
$x + 5x + 4x = 180^\circ$
$10x = 180^\circ$
$\text{x}=\frac{180^\circ}{10}$
$x = 18^\circ .$

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MCQ 2081 Mark

Two straight lines $AB$ and $CD$ cut each other at $O$. If $\angle\text{BOD}=63^\circ,$ then $\angle\text{BOC}=$

A
$153^\circ$
B
$17^\circ$
✓
$117^\circ$
D
$63^\circ$

Answer

Correct option: C.

$117^\circ$

$\angle\text{BOD}+\angle\text{BOC}=180^\circ$ (Linear pair)
$63^\circ+\angle\text{BOC}=180^\circ$
$\text{BOC}=117^\circ.$

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MCQ 2091 Mark

In Fig, $PQ \| RS$, $\angle\text{AEF}=95^\circ,\angle\text{BHS}=110^\circ,$ and $\angle\text{ABC}=\text{x}^\circ.$ Then the value of $x$ is:

A
$70^\circ$
B
$15^\circ$
✓
$25^\circ$
D
$35^\circ$

Answer

Correct option: C.

$25^\circ$

Given that,
$PQ \| RS$
$\angle\text{AEF}=95^\circ$
$\angle\text{BHS}=110^\circ$
$\angle\text{ABC}=\text{x}^\circ$
$\angle\text{AEF}=\angle\text{AGH}=95^\circ$ (Corresponding angles)
$\angle\text{AGH}+\angle\text{HGB}=180^\circ$ (Linear pair)
$95^\circ+\angle\text{HGB}=180^\circ$
$\angle\text{HGB}=85^\circ$
$\angle\text{BHS}+\angle\text{BHG}=180^\circ $(Linear pair)
$110^\circ+\angle\text{BHG}=180^\circ$
$\angle\text{BHG}=70^\circ$
In $\triangle\text{BHG},$
$\angle\text{BHG}+\angle\text{HGB}+\angle\text{GBH}=180^\circ$
$70^\circ+85^\circ+\angle\text{GBH}=180^\circ$
$\angle\text{GBH}=25^\circ$
Thus,
$\angle\text{ABC}=\angle\text{GBH}=25^\circ.$

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MCQ 2101 Mark

Two lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ,$then $\angle\text{AOC}=$

A
$70^\circ$
B
$80^\circ$
✓
$90^\circ$
D
$180^\circ$

Answer

Correct option: C.

$90^\circ$

$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ$ [Given]
From figure,
$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}+\angle\text{DOA}=360^\circ$
$\Rightarrow\ 270^\circ+\angle\text{DOA}=360^\circ$
$\Rightarrow\ \angle\text{DOA}=360^\circ-270^\circ=90^\circ$
Now,
$\angle\text{DOA}+\angle\text{AOC}=180^\circ$
$\Rightarrow\ \angle\text{AOC}=180^\circ-90^\circ=90^\circ$

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MCQ 2111 Mark

In the given figure, $\angle\text{OAB}=75^\circ, \angle\text{OBA}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ then, $\angle\text{ODC}=?$

A
$20^\circ$
B
$25^\circ$
✓
$30^\circ$
D
$35^\circ$

Answer

Correct option: C.

$30^\circ$

In $\triangle\text{OAB},$ we have
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ (Angle sum property)
$\Rightarrow55^\circ+75^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=50^\circ$
$\Rightarrow\angle\text{COD}=\angle\text{AOB}=50^\circ$ (Vertivcally opposite angles)
In $\triangle\text{OCD},$ we have
$\angle\text{COD}+\angle\text{OCD}+\angle\text{ODC}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+100^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$

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MCQ 2121 Mark

In the adjoining figure, if $A = 40^\circ , B = 95^\circ $ and $D = 60^\circ $, then $E$ is equal to:

✓
$75^\circ$
B
$55^\circ$
C
$65^\circ$
D
$85^\circ$

Answer

Correct option: A.

$75^\circ$

in $\triangle\text{ABC}$$\angle\text{ABC}+\angle\text{BCA}+\angle\text{CAB}=180^\circ$ (Angle sum property)
$\angle\text{BCA}=180^\circ-95^\circ-40^\circ=45^\circ$
$\angle\text{BCA}=\angle\text{ECD}=45^\circ$ (Vertically opposite angle)
In $\triangle\text{ECD}$
$\angle\text{ECD}+\angle\text{DEC}+\angle\text{CDE}=180^\circ$ (Angle sum property)
$\angle\text{ECD}=180^\circ-45^\circ-60^\circ=75^\circ.$

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MCQ 2131 Mark

In the given figure, the value of x which makes $POQ$ a straight line is:

A
$30^\circ$
✓
$25^\circ$
C
$35^\circ$
D
$40^\circ$

Answer

Correct option: B.

$25^\circ$

We know that he measure of a straight angle is $180^\circ$
$(2x + 30^\circ ) + 4x = 180^\circ$
$2x + 30^\circ + 4x = 180^\circ$
$6x = 180^\circ - 30^\circ$
$6x = 150^\circ$
$\text{x}=\frac{150^\circ}{6}=25^\circ.$

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MCQ 2141 Mark

In figure, if $AB \| CD$, then the value of $x$ is:

A
$20^\circ$
✓
$30^\circ$
C
$45^\circ$
D
$60^\circ$

Answer

Correct option: B.

$30^\circ$

From figure,
$\angle\text{DPQ}+\angle\text{x}^\circ=180^\circ\dots(1)$ [linear pair]
Also,
$\angle\text{DPQ}=\angle\text{AQP}$ [Interior opposite angles]
$\Rightarrow\ \angle\text{DPQ}=120^\circ+\text{x}$
From $(1),$
$120^\circ+\text{x}+\text{x}=180^\circ$
$\Rightarrow\ 2\text{x}=60^\circ$
$\Rightarrow\text{x}=30^\circ$

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MCQ 2151 Mark

In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=4\text{x}^\circ$ and $\angle\text{BOC}=5\text{x}^\circ$ then $\angle\text{AOC}=?$

A
$40^\circ$
B
$60^\circ$
✓
$80^\circ$
D
$100^\circ$

Answer

Correct option: C.

$80^\circ$

Since $AOB$ is a straight line,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow4\text{x}+5\text{x}=180^\circ$
$\Rightarrow9\text{x}=180^\circ$
$\Rightarrow\text{x}=206^\circ$
So, $\angle\text{AOC}=4\text{x}=4(20)=80^\circ$

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MCQ 2161 Mark

In a figure, if $OP \| RS$, $\angle\text{OPQ}=110^\circ$ and $\angle\text{QRS}=130^\circ,$ then $\angle\text{PQR}$ is equal to:

✓
$60^\circ$
B
$70^\circ$
C
$40^\circ$
D
$50^\circ$

Answer

Correct option: A.

$60^\circ$

Produce $OP$ to intersect $RQ$ at point $N.$
Now, $OP \| RS$ and transversal RN intersects them at N and R respectively.
$\therefore\angle\text{RNP}=\angle\text{SRN}$ (Alternate interior angles)
$\Rightarrow\angle\text{RNP}=130^\circ$
$\therefore\angle\text{PNQ}=180^\circ−130^\circ=50^\circ$ (Linear pair)
$\angle\text{OPQ}=\angle\text{PNQ}+\angle\text{PQN}$ (Exterior angle property)
$\Rightarrow110^\circ= 50^\circ+\angle\text{PQN}$
$\Rightarrow\angle\text{PQN}=110^\circ-50^\circ=60^\circ=\angle\text{PQR}.$

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MCQ 2171 Mark

In the below figure, the value of $x$ is:

A
$30^\circ$
✓
$10^\circ$
C
$15^\circ$
D
$25^\circ$

Answer

Correct option: B.

$10^\circ$

$2x - 20^\circ + x + 100^\circ = 110^\circ $ (Exterior angle is equal to sum of its interior opposite angles)$3x = 110^\circ - 100^\circ + 20^\circ$
$x = 10^\circ .$

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MCQ 2181 Mark

In the given figure, $\angle\text{BAC}=40^\circ,\angle\text{ACB}=90^\circ$ and $\angle\text{BED}=100^\circ,$ Then $\angle\text{BDE}=?$

A
$50^\circ$
B
$40^\circ$
C
$25^\circ$
✓
$30^\circ$

Answer

Correct option: D.

$30^\circ$

In $\triangle\text{ABC}$$\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$\angle\text{ABC}=180^\circ-90^\circ-40^\circ$
$\angle\text{ABC}=50^\circ$
In $\angle\text{BED}$
$\angle\text{BED}+\angle\text{EBD}+\angle\text{BDE}=180^\circ$ (Angle sum property)
$\angle\text{BDE}=180^\circ-50^\circ-100^\circ$
$\angle\text{BDE}=30^\circ.$

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MCQ 2191 Mark

In figure, if $l \| m$, what is the value of $x?$

✓
$60$
B
$50$
C
$45$
D
$30$

Answer

Correct option: A.

$60$

$3y^\circ = 2y^\circ + 25^\circ $ [Alternate angles]
$\Rightarrow y^\circ = 25^\circ $
Now,
$x^\circ + 15^\circ = 2y^\circ + 25^\circ $ [Opposite angles]
$\Rightarrow x = 2y^\circ + 25^\circ - 15^\circ $
$\Rightarrow x = 2y^\circ + 10^\circ $
$\Rightarrow x = 2 \times 25^\circ + 10^\circ $
$\Rightarrow x = 60^\circ $

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MCQ 2201 Mark

Which of the following statements is false?

✓
Through a given point, only one straight line can be drawn.
B
Through two given points, it is possible to draw one and only one straight line.
C
Two straight lines can intersect only at one point.
D
A line segment can be produced to any desired length.

Answer

Correct option: A.

Through a given point, only one straight line can be drawn.

Option $(a)$ is false, since through a given point we can draw an infinite number of straight lines.

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MCQ 2211 Mark

In figure, $\text{PQ }||\text{ RS},\angle\text{QPR}=70^\circ,\angle\text{ROT}=20^\circ$ find the value of $x.$

A
$20^\circ$
✓
$50^\circ$
C
$70^\circ$
D
$110^\circ$

Answer

Correct option: B.

$50^\circ$

$\text{PQ }||\text{ RS},\angle\text{QPR}=\angle\text{SRO}=70^\circ$ (Corresponding, Angle)
Now in $\triangle\text{RTO}$
$x + 20^\circ = 70^\circ $ (exterior angle)
$x = 70^\circ - 20^\circ$
$x = 50^\circ .$

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MCQ 2221 Mark

The complement of $(90-a)^0$ is:

A
$(90+2 a)^{\circ}$
B
$(90-a)^0$
✓
$a^{\circ}$
D
$-a^{\circ}$

Answer

Correct option: C.

$a^{\circ}$

We know that the sum of complementary angles are $90^{\circ}$
Let the complementary angle of $(90-\mathrm{a})^{\circ}$ be $x$
$x+(90-a)^{\circ}=90^{\circ}$
$x=90^{\circ}-(90-a)^{\circ}$
$x=90^{\circ}-90^{\circ}+a^{\circ}$
$x=a^{\circ}$.

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MCQ 2231 Mark

In Fig. if lines $l$ and $m$ are parallel, then $x =$

A
$85^{\circ}$
B
$65^{\circ}$
C
$20^{\circ}$
✓
$45^{\circ}$

Answer

Correct option: D.

$45^{\circ}$

$l \| m$ Let transversal be n and $\angle1=65^\circ$
$\angle2=20^\circ$
$\angle3=\text{x}$
Since,
$l \| m$ and n cuts them so,
$\angle1+\angle4=180^\circ$ (Co. interior angle)
$65^\circ+\angle4=80^\circ$
$\angle4=115^\circ\text{(i)}$
$\angle4=\angle5=115^\circ$ (Vertically opposite angle)
$\angle2+\angle5+\angle3=180^\circ$
$20^{\circ} + 115^{\circ}+ x = 180^{\circ}$
$x = 45^{\circ}$.

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MCQ 2241 Mark

In the given figure $AB$ is a mirror, $PQ$ is the incident ray and $QR$ is the reflected ray. If $\angle\text{PQR}=108^\circ$ then $\angle\text{AQP}=?$

A
$18^\circ$
B
$54^\circ$
C
$72^\circ$
✓
$36^\circ$

Answer

Correct option: D.

$36^\circ$

According to question,
$\angle\text{AQP}=\angle\text{BQR}=\text{x}$
$\angle\text{AQP}+\angle\text{BQR}+\angle\text{PQR}=180^\circ$ (Linear Pair)
$2x + 108^\circ = 180^\circ$
$x = 36^\circ $.

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MCQ 2251 Mark

In the adjoining figure, if $m \| n$, then $\angle4+\angle7$ is equal to:

✓
$180^\circ$
B
$120^\circ$
C
$90^\circ$
D
$150^\circ$

Answer

Correct option: A.

$180^\circ$

$\angle1=\angle5$ (Corresponding angle)
$\angle7=\angle5$ (Vertically Opposite angle)
$\angle4+\angle1=180^\circ$ (Linear Pair)
$\angle4+\angle7=180^\circ$ (From above equations).

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MCQ 2261 Mark

In the given figure, $AB \| CD$. If $\angle\text{EAB}=50^\circ$ and $\angle\text{ECD}=60^\circ$ then $\angle\text{AEB}=?$

A
$50^\circ$
B
$60^\circ$
✓
$70^\circ$
D
$55^\circ$

Answer

Correct option: C.

$70^\circ$

Let $\angle\text{AEB}=\text{x}^\circ$
Now, $AB \| CD$ and $BC$ is the transversal
$\therefore\angle\text{ABE}=\angle\text{BCD}=60^\circ$ (Alterante angles)
In $\triangle\text{ABE},$
$\angle\text{BAE}+\angle\text{AEB}+\angle\text{ABE}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+\text{x}^\circ+60^\circ=180^\circ$
$\Rightarrow\text{x}=70^\circ$
$\therefore\angle\text{AEB}=70^\circ$

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MCQ 2271 Mark

In the given figure, $AB \| CD$. If $\angle\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ$ then $\angle\text{OCD}=?$

✓
$130^\circ$
B
$150^\circ$
C
$80^\circ$
D
$100^\circ$

Answer

Correct option: A.

$130^\circ$

Construction: Through $O$, draw $OE \| AB \| CD$
$\Rightarrow\angle\text{BAO}+\angle\text{EOA}=180^\circ$
$\Rightarrow100^\circ+\angle\text{EOA}=180^\circ$
$\Rightarrow\angle\text{EOA}=80^\circ$
So, $\angle\text{EOC}=\angle\text{EAO}-\angle\text{COA}=80^\circ-30^\circ=50^\circ$
Since $CD \| EO$
$\angle\text{OCD}+\angle\text{EOC}=1806^\circ$
$\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
$\Rightarrow\angle\text{OCD}=130^\circ$

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