M.C.Q

Question 1011 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 $\angle\text{COE}:$

Answer

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

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Question 1021 Mark

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

Answer

100º
Solution:

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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Question 1031 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 $\angle\text{BOD}=?$

Answer

285º
Solution:

$\angle\text{ABO}+\angle\text{BOE}=180^\circ$ (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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Question 1041 Mark

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

Answer

A right triangle.
Solution:
In a right triangle, one angle is 90° and the sum of acute angles of a right triangle is 90°.

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Question 1051 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:

Answer

$50^\circ$
Solution:
$\angle\text{BEC}+\angle\text{EBC}=\angle\text{ECD}$ (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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Question 1061 Mark

In the given figure x = 30º, the value of Y is:

Answer

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

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Question 1071 Mark

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

Answer

115º
Solution:
$\angle\text{AOC}+\angle\text{BOD}=1306^\circ$ (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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Question 1081 Mark

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

Answer

80º
Solution:
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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Question 1091 Mark

Measurement of reflex angle is:

Answer

Between 180º and 360º
Solution:
Let x be the angle
then its reflex angle is 360º - x
and in any triangle the angle lies between 0 to 180º

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Question 1101 Mark

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

Answer

105°
Solution:

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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Question 1111 Mark

In figure, if $l_1 || l_{2,}$what is the value of $x$?

Answer

From figure,
$\angle\text{ERC}=\angle\text{RPA} [$Corresponding angles are equal$]$
$\Rightarrow \angle\text{ERC}=37^\circ=\angle\text{RPA}$
Also,
$\angle\text{RPA}=\angle\text{BPF} [$Opposite angles$]$
$\Rightarrow\ \angle\text{RPA}=37^\circ=\angle\text{BPF}$
Now,
$\angle\text{QPB}+\angle\text{BPF}+\angle\text{FPG}=180^\circ$
$\Rightarrow\ \text{x}^\circ+37^\circ+58^\circ=180^\circ$
$\Rightarrow\ \text{x}^\circ=85^\circ$

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Question 1121 Mark

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

Answer

An acute angle.
Solution:
If two angles are complements of each other, that is, the sum of their measures is 90º, then each angle is an acute angle.

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Question 1131 Mark

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

Answer

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

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Question 1141 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.

Answer

54º
Solution:
Let the measure of the required angle be xº
Then, the measure of its complement will be (90 - x)º
Therefore, 2x = 3 (90 - x)
⇒ 2x = 270 - 3x
⇒ 5x = 270
⇒ x = 54º.

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Question 1151 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:

Answer

108°
Solution:

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 $\angle\text{BRS}=72^\circ$
⇒ Larger angle is $\angle\text{DSR}.$

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Question 1161 Mark

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

Answer

45°
Solution:

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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Question 1171 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.

Answer

155°
Solution:
$$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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Question 1181 Mark

In figure, if $l_1 || l_2$ and $l_3 || l_4,$ what is $y$ in terms of $x$?

Answer

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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Question 1191 Mark

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

Answer

40°
Solution:

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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Question 1201 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,

Answer

60°
Solution:
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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Question 1211 Mark

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

Answer

135º
Solution:
Let the required angle be x
Supplement = 180º - x
According to question,
x = 3 (180º - x)
x = 540º - 3x
x = 135º.

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Question 1221 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:

Answer

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

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Question 1231 Mark

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

Answer

(ii) and (iii) only
Solution:

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,
$\angle\text{a}^\circ+\angle\text{e}^\circ+\angle\text{c}^\circ=180^\circ$
⇒ (ii) is correct
Now,
$\angle\text{FOB}\neq\angle\text{DOB}$
$\Rightarrow\ \angle\text{a}^\circ+\angle\text{b}^\circ\neq\angle\text{d}^\circ+\angle\text{c}^\circ$
⇒ (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,
$\angle\text{b}^\circ=\angle\text{f}^\circ=\angle\text{e}^\circ+\angle\text{c}^\circ$
⇒ (iii) is correct.

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Question 1241 Mark

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

Answer

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

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Question 1251 Mark

In figure, the value of x, is:

Answer

20
Solution:

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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Question 1261 Mark

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

Answer

90º
Solution:
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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Question 1271 Mark

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

Answer

20
Solution:
Since, AB || CD
And, BD cuts them
y + 2y + y + 5y = 180º (Consecutive interior angle)
9y = 180º
y = 20º.

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Question 1281 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 $\phi=?$

Answer

45º
Solution:
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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Question 1291 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,

Answer

40°
Solution:
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°]
$\Rightarrow2\text{x}+4\text{x}+3\text{x}=180^\circ\Rightarrow9\text{x}=180^\circ\Rightarrow\text{x}=\frac{180^\circ}{9}=20^\circ$
$\therefore\ \angle\text{A}=2\text{x}=2\times20^\circ=40^\circ$
$\angle\text{B}=4\text{x}=4\times20^\circ=80^\circ$
And $\angle\text{C}=3\text{x}=3\times20^\circ=60^\circ$
Hence, the smallest angles of a triangle is 40° and option (b) is correct answer.

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Question 1301 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 $\angle\text{BOC}=?$

Answer

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

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Question 1311 Mark

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

Answer

145°
Solution:

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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Question 1321 Mark

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

Answer

55º
Solution:
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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Question 1331 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:

Answer

25°
Solution:

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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Question 1341 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 $\angle\text{A}=?$

Answer

75°
Solution:
$\angle\text{ABD}+\angle\text{ABC}=180^\circ$(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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Question 1351 Mark

Each angle of an equilateral triangle is:

Answer

60º
Solution:
Let the angle of an equilateral triangle be xo
x + x + x = 180º (Angle sum property)
3x = 180º
x = 60º.

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Question 1361 Mark

Which of the following pairs of angles are complementary?

Answer

25º, 65º
Solution:
Complementary angles always add up to 90º
25º + 65º = 90º
Therefore this is the right option (by substitution).

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Question 1371 Mark

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

Answer

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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Question 1381 Mark

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

Answer

75º
Solution:
Let the measure of the angle be xº,
So, its complement = (90 - x)º
According to the given condition,
x = 5(90 - x)
⇒ x = 450 - 5x
⇒ 6x = 450
⇒ x = 75º
So, the angle measures 75º.

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Question 1391 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.

Answer

145°
Solution:
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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Question 1401 Mark

Measurement of reflex angle is:

Answer

Between 180º and 360º
Solution:
Let x be the angle then its reflex angle is 360º - x and in any triangle, the angle lies between 0 to 180º.

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Question 1411 Mark

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

Answer

80º, 65º, 35º
Solution:
Since ABC is a triangle
A + B + C = 180º (Angle sum property) (i)
A - B = 15º
A = 15º + B (ii)
B - C = 30º
C = B - 30º (iii)
From (i) equation
15º + B + B + B - 30º = 180º
B = 65º
From equation (ii) and (iii)
A = 15º + B = 15º + 65º = 80º
C = B - 30º = 65º - 30º = 35º.

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Question 1421 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:

Answer

60º
Solution:
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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Question 1431 Mark

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

Answer

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

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Question 1441 Mark

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

Answer

115º
Solution:
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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Question 1451 Mark

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

Answer

128º
Solution:

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

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Question 1461 Mark

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

Answer

80º
Solution:
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)
⇒ 2k + 3k + 4k = 180º
⇒ 9k = 180º
⇒ k = 20º.

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Question 1471 Mark

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

Answer

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

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Question 1481 Mark

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

Answer

8
Solution:

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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Question 1491 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 $\angle\text{AOD}=$

Answer

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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Question 1501 Mark

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

Answer

75º
Solution:
Let the measure of the required angle be xº
Then, the measure of its complement will be (90 − x)º
Therefore, x = 5 (90 - x)
⇒ x = 450 - 5x
⇒ 6x = 450
⇒ x = 75º.

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MATHS M.C.Q