M.C.Q

Question 511 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:

Answer

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

View full question & answer→

Question 521 Mark

If two angles of a triangle are 30º and 45º, what is the measure of the third angle?

Answer

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

View full question & answer→

Question 531 Mark

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

Answer

100º
Solution:

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

View full question & answer→

Question 541 Mark

Given that lines $l_1, l_2$ and $l_3$ in figure are parallel. The value of $x$ is:

Answer

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 $\angle\text{x}=140^\circ.$

View full question & answer→

Question 551 Mark

In figure, if $l_1 \ \| \ l_2,$ what isb the value of $y$ ?

Answer

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

View full question & answer→

Question 561 Mark

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

Answer

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

View full question & answer→

Question 571 Mark

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

Answer

Alternate exterior angles.
Solution:
$\angle\text{a}$ and $\angle\text{g}$ are on alternate side and are exterior.

View full question & answer→

Question 581 Mark

Two planes intersect each other to form a:

Answer

Straight line.
Solution:

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

View full question & answer→

Question 591 Mark

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

Answer

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$)$
$\text{x}+90^\circ+55^\circ=180^\circ$
$\text{x}=35^\circ.$

View full question & answer→

Question 601 Mark

The incorrect statement is:

Two lines drawn in a plane always intersect at a point.
A line segment has definite length.
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

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

View full question & answer→

Question 611 Mark

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

Answer

130°
Solution:

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,
$\text{x}=\angle\text{ACP}=180^\circ-\angle\text{ACE}$
$=180^\circ-50^\circ=130^\circ$

View full question & answer→

Question 621 Mark

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

Answer

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
$2x = 130^\circ$
Hence $x = 65^\circ$
Substituting this value of $x$ in either one of the equations $(1)$ or $(2)$
We get $y = 25^\circ .$

View full question & answer→

Question 631 Mark

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

Answer

126º
Solution:
AB || CD
x + y = 180º and y = p (Vertically opposite angles)
Also, CD || EF and t is the transversal.
$\therefore$ p + z = 180º
⇒ y + z = 180º $(\because\text{p}=\text{y})$
$\therefore$ x + y = y + z
⇒ 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 = 126º $(\because\text{x}=\text{z})$

View full question & answer→

Question 641 Mark

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

Answer

108º
Solution:
$\angle\text{a}+\angle\text{b}=180^\circ$ (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).

View full question & answer→

Question 651 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 =?

Answer

25º
Solution:
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)
$90^\circ+\angle\text{MAN}=115^\circ$
$\angle\text{MAN}=25^\circ.$

View full question & answer→

Question 661 Mark

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

Answer

30º
Solution:
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 $\triangle\text{ODC}$
$\angle\text{ODC}+\angle\text{DOC}+\angle\text{DCO}=180^\circ$
$\angle\text{ODC}=180^\circ-100^\circ-50^\circ$
$\angle\text{ODC}=30^\circ$

View full question & answer→

Question 671 Mark

In Fig., which of the following statements must be true?

a + b = d + c
a + c + e = 180º
b + f = c + e

Answer

(ii) and (iii) only
Solution:
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 = 180º
a + c + e = 180º [From (ii)].

View full question & answer→

Question 681 Mark

When two straight lines intersect:

Adjacent angles are complementary
Adjacent angles are supplementary.
Opposite angles are equal.
Opposite angles are supplementary.

Of these statements:

Answer

(ii) and (iii) are correct
Solution:
When two straight lines intersect them, Adjacent angles are supplementary and opposite angles are equal.

View full question & answer→

Question 691 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:
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 $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}=\angle\text{ACD}$
$\angle\text{ABC}+2\angle\text{EBC}=2\angle\text{ECD}$
$\angle\text{ABC}=2(\angle\text{ECD}-\angle\text{EBC})$
$\angle\text{ABC}=2(\angle(\text{BEC}))$
$\angle\text{ABC}=50^\circ$

View full question & answer→

Question 701 Mark

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

Answer

30°
Solution:

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),
$120^\circ+\text{x}+\text{x}=180^\circ$
$\Rightarrow\ 2\text{x}=60^\circ$
$\Rightarrow\text{x}=30^\circ$

View full question & answer→

Question 711 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 $\phi=?$

Answer

45º
Solution:
$\angle\text{AOD}=\angle\text{COB}=\theta$
$\angle\text{AOC}=\angle\text{BOD}=\phi$
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, $3\phi+\phi=180$
$\Rightarrow4\phi=180$
$\Rightarrow\phi=45$

View full question & answer→

Question 721 Mark

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

Answer

60º
Solution:
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)

View full question & answer→

Question 731 Mark

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

Answer

35°
Solution:

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

View full question & answer→

Question 741 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:

Answer

36°
Solution:
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°.

View full question & answer→

Question 751 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:-

Answer

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

View full question & answer→

Question 761 Mark

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

Answer

145º
Solution:
Given, AB || CD || EF and GH || KL
Produce 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, $\angle\text{HKL}=\angle3+\angle\text{MKL}$
= 25º + 120º
= 145º.

View full question & answer→

Question 771 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:

Answer

108º
Solution:
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 = 180º
2x + 3x = 180º
5x = 180º
$\text{x} = \frac{180^\circ}{5} = 36^\circ$
x = 36º
3x = 3 × 36º = 108º.

View full question & answer→

Question 781 Mark

If one of the angles of a triangle is 130º then the angle between the bisectors of the other two angles can be:

Answer

155º
Solution:

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 $\triangle\text{BOC},$
$\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\angle\text{BOC}=180^\circ$
$\Rightarrow25^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=155^\circ$

View full question & answer→

Question 791 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, $\angle\text{B}?$

Answer

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

View full question & answer→

Question 801 Mark

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

Answer

55º
Solution:

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)
$\Rightarrow120^\circ=70^\circ+\angle\text{QPR}$
$\Rightarrow\angle\text{QPR}=506^\circ$

View full question & answer→

Question 811 Mark

In Fig., the value of x is:

Answer

20º
Solution:
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 + 10º + 90º = 180º
4x + 100º = 180º
4x = 80º
x = 20º.

View full question & answer→

Question 821 Mark

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

Answer

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

View full question & answer→

Question 831 Mark

Which of the following statements is false?

Answer

Through a given point, only one straight line can be drawn.
Solution:
This statement is false because we can draw infinitely many straight lines through a given point.

View full question & answer→

Question 841 Mark

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

Answer

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

View full question & answer→

Question 851 Mark

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

Answer

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

View full question & answer→

Question 861 Mark

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

Answer

Infinity.
Solution:

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".

View full question & answer→

Question 871 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:

Answer

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

View full question & answer→

Question 881 Mark

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

Answer

80º
Solution:
By angle sum property,
2x + 3x + 4x = 180°
⇒ 9x = 180°
⇒ x = 20°
Hence, largest angle = 4x = 4(20°) = 80°

View full question & answer→

Question 891 Mark

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

Answer

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

View full question & answer→

Question 901 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:

Answer

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

View full question & answer→

Question 911 Mark

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

Answer

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
$x + y = 180^\circ - w + z.$

View full question & answer→

Question 921 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

90°
Solution:

$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ$ [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,
$\angle\text{DOA}+\angle\text{AOC}=180^\circ$
$\Rightarrow\ \angle\text{AOC}=180^\circ-90^\circ=90^\circ$

View full question & answer→

Question 931 Mark

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

Answer

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

View full question & answer→

Question 941 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 $\angle\text{AQP}=?$

Answer

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

View full question & answer→

Question 951 Mark

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

Answer

18°
Solution:

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
$\angle\text{x}^\circ+\angle5\text{x}^\circ+\angle4\text{x}=180^\circ$
$\Rightarrow\ \angle10\text{x}^\circ=180^\circ$
$\Rightarrow\ \angle\text{x}^\circ=18^\circ$

View full question & answer→

Question 961 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:
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.

View full question & answer→

Question 971 Mark

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

Answer

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

View full question & answer→

Question 981 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:

Answer

145º
Solution:
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]
$=25^\circ+180^\circ-60^\circ=205^\circ- 60^\circ=145^\circ.$

View full question & answer→

Question 991 Mark

In the given x =?

Answer

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

View full question & answer→

Question 1001 Mark

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

Answer

50º
Solution:
$\angle\text{APQ}=\angle\text{PQR}=70^\circ$ (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)
$\angle\text{QPR}=180^\circ-70^\circ-60^\circ=50^\circ.$

View full question & answer→

MATHS M.C.Q