CBSE Class 9 Maths Important Questions - Chapter 6 Lines and Angles

1-Marks:

1. Measurement of reflex angle is

(i) ${90}^{\circ }$

(ii) between $0^{\circ }$ and ${90}^{\circ }$

(iii) between ${90}^{\circ }$ and ${180}^{\circ }$

(iv) between ${180}^{\circ }$ and ${360}^{\circ }$

Ans: (iv) between ${180}^{\circ }$ and ${360}^{\circ }$

2. The sum of angle of a triangle is

(i) $0^{\circ }$

(ii) ${90}^{\circ }$

(iii) ${180}^{\circ }$

(iv) none of these

Ans: (iii) ${180}^{\circ }$

3. In fig if $\text{ x}={30}^{\circ }$ then y = 

(i) ${90}^{\circ }$

(ii) ${180}^{\circ }$

(iii) ${150}^{\circ }$

(iv) ${210}^{\circ }$

Ans: (iii) ${150}^{\circ }$

4. If two lines intersect each other then

(i) Vertically opposite angles are equal

(ii) Corresponding angle are equal

(iii) Alternate interior angle are equal

(iv) None of these

Ans: (i) Vertically opposite angles are equal

5. The measure of Complementary angle of ${63}^{\circ }$ is

(a) ${30}^{\circ }$

(b) ${36}^{\circ }$

(c) ${27}^{\circ }$

(d) None of there

Ans: (c) ${27}^{\circ }$

6. If two angles of a triangle is ${30}^{\circ }$ and ${45}^{\circ }$ what is measure of third angle

(a) ${95}^{\circ }$

(b) ${90}^{\circ }$

(c) ${60}^{\circ }$

(d) ${105}^{\circ }$

Ans: (d) ${105}^{\circ }$

7. The measurement of Complete angle is

(a) $0^{\circ }$

(b) ${90}^{\circ }$

(c) ${180}^{\circ }$

(d) ${360}^{\circ }$

Ans: (d) ${360}^{\circ }$        

8. The measurement of sum of linear pair is

(a) ${180}^{\circ }$

(b) ${90}^{\circ }$

(c) ${270}^{\circ }$

(d) ${360}^{\circ }$

Ans: (a) ${180}^{\circ }$

9. The difference of two complementary angles is ${40}^{\circ }$. The angles are

(a) ${65}^{\circ },{35}^{\circ }$

(b) ${70}^{\circ },{30}^{\circ }$

(c) ${25}^{\circ },{65}^{\circ }$

(d) ${70}^{\circ },{110}^{\circ }$

Ans: (c) ${25}^{\circ },{65}^{\circ }$

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

(a) In the interior of $\mathrm{\angle }ABC$

(b) In the interior of $\mathrm{\angle }ABC$

(c) On the $\mathrm{\angle }ABC$

(d) On the both sides of $\overrightarrow{BA}$

Ans: (c) On the $\mathrm{\angle }ABC$

11. The complement of $(90-a{)}^0$ is

(a) $-a^0$

(b) $(90+2a{)}^0$

(c) $(90-a{)}^0$

(d) $a^0$

Ans: (d) $a^0$

12. The number of angles formed by a transversal with a pair of lines is

(a) 6

(b) 3

(c) 8

(d) 4

Ans: (c) 8

13. In fig $$L_1\parallel L_2$$ and $\mathrm{\angle }1={52}^{\circ }$ the measure of $\mathrm{\angle }2$ is.

(A) ${38}^{\circ }$

(B) ${128}^{\circ }$

(C) ${52}^{\circ }$

(D) ${48}^{\circ }$

Ans: (B) ${128}^{\circ }$

14. In $fig\text{ x}={30}^{\circ }$ the value of Y is 

(A) ${10}^{\circ }$

(B) ${40}^{\circ }$

(C) ${36}^{\circ }$

(D) ${45}^{\circ }$

Ans: (B) ${40}^{\circ }$

15. Which of the following pairs of angles are complementary angle?

(A) ${25}^{\circ },{65}^{\circ }$

(B) ${70}^{\circ },{110}^{\circ }$

(C) ${30}^{\circ },{70}^{\circ }$

(D) ${32.1}^{\circ },{47.9}^{\circ }$

Ans: (A) ${25}^{\circ },{65}^{\circ }$

16. In fig the measures of $\mathrm{\angle }1$ is.

(A) ${158}^{\circ }$

(B) ${138}^{\circ }$

(C) ${42}^{\circ }$

(D) ${48}^{\circ }$

Ans: (C) ${42}^{\circ }$

17. In figure the measure of $\mathrm{\angle }a$ is

(a) ${30}^{\circ }$

(b) ${150}^0$

(c) ${15}^{\circ }$

(d) ${50}^{\circ }$

Ans: (a) ${30}^{\circ }$

18. The correct statement is-

F point in common.

Ans: (c) Three points are collinear if all of them lie on a line.

19. One angle is five times its supplement. The angles are-

(a) ${15}^{\circ },{75}^{\circ }$

(b) ${30}^{\circ },{150}^{\circ }$

(c) ${36}^{\circ },{144}^0$

(d) ${160}^{\circ },{40}^{\circ }$

Ans: (b) ${30}^{\circ },{150}^{\circ }$

20. In figure if  and $\mathrm{\angle }1:\mathrm{\angle }2=1:2.$ the measure of $\mathrm{\angle }8$ is

(a) ${120}^{\circ }$

(b) ${60}^{\circ }$

(c) ${30}^{\circ }$

(d) ${45}^{\circ }$

Ans:  (b) ${60}^{\circ }$

2-Marks:

1. In Fig. 6.13, lines $\text{AB}$ and $\text{CD}$ intersect at $O.$ If $\mathrm{\angle }\text{AOC}+\mathrm{\angle }\text{BOE}={70}^{\circ }$ and $\mathrm{\angle }\text{BOD}={40}^{\circ }$, find $\mathrm{\angle }\text{BOE}$ and reflex $\mathrm{\angle }\text{COE}.$

Ans: According to the question given that, $\mathrm{\angle }AOC+\mathrm{\angle }BOE={70}^{\circ }$ and $\mathrm{\angle }BOD={40}^{\circ }$.

We need to find $\mathrm{\angle }BOE$ and reflex $\mathrm{\angle }COE$.

According to the given figure, we can conclude that $\mathrm{\angle }COB$ and $\mathrm{\angle }AOC$ form a linear pair.

As we also know that sum of the angles of a linear pair is ${180}^{\circ }$.

So, $\mathrm{\angle }COB+\mathrm{\angle }AOC={180}^{\circ }$

Because,$\mathrm{\angle }COB=\mathrm{\angle }COE+\mathrm{\angle }BOE$, or

So, $\mathrm{\angle }AOC+\mathrm{\angle }BOE+\mathrm{\angle }COE={180}^{\circ }$

$\Rightarrow {70}^{\circ }+\mathrm{\angle }COE={180}^{\circ }$

$\Rightarrow \mathrm{\angle }COE={180}^{\circ }-{70}^{\circ }$

$={110}^{\circ }.$

Reflex $\mathrm{\angle }COE={360}^{\circ }-\mathrm{\angle }COE$

$={360}^{\circ }-{110}^{\circ }$

$={250}^{\circ }.$

$\mathrm{\angle }AOC=\mathrm{\angle }BOD$ (Vertically opposite angles), or

$\mathrm{\angle }BOD+\mathrm{\angle }BOE={70}^{\circ }$

But, according to the question given that $\mathrm{\angle }BOD={40}^{\circ }$.

${40}^{\circ }+\mathrm{\angle }BOE={70}^{\circ }$

$\mathrm{\angle }BOE={70}^{\circ }-{40}^{\circ }$

$={30}^{\circ }$.

Hence, we can conclude that Reflex $\mathrm{\angle }COE={250}^{\circ }$ and $\mathrm{\angle }BOE={30}^{\circ }$.

2. In the given figure, $\mathrm{\angle }PQR=\mathrm{\angle }PRQ$, then prove that $\mathrm{\angle }PQS=\mathrm{\angle }PRT$.

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Ans: According to the question we need to prove that $\mathrm{\angle }PQS=\mathrm{\angle }PRT$.

According to the question given that $\mathrm{\angle }PQR=\mathrm{\angle }PRQ$.

According to the given figure, we can conclude that $\mathrm{\angle }PQS$ and $\mathrm{\angle }PQR$, and $\mathrm{\angle }PRS$ and $\mathrm{\angle }PRT$ form a linear pair.

As we also know that sum of the angles of a linear pair is ${180}^{\circ }$.

So, $\mathrm{\angle }PQS+\mathrm{\angle }PQR={180}^{\circ }$, and(i)

$\mathrm{\angle }PRQ+\mathrm{\angle }PRT={180}^{\circ }$..(ii)

According to the equations (i) and (ii), we can conclude that

$\mathrm{\angle }PQS+\mathrm{\angle }PQR=\mathrm{\angle }PRQ+\mathrm{\angle }PRT$

But, $\mathrm{\angle }PQR=\mathrm{\angle }PRQ.$

So, $\mathrm{\angle }PQS=\mathrm{\angle }PRT$

Hence, the desired result is proved.

3. In the given figure, find the values of $\text{x}$ and $\text{y}$ and then show that .

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Ans: According to the question we need to find the value of $x$ and $y$ in the figure given below and then prove that 

According to the figure, we can conclude that $y={130}^{\circ }$ (Vertically opposite angles), and

$x$ and ${50}^{\circ }$ form a pair of linear pair.

As we also know that the sum of linear pair of angles is ${180}^{\circ }$.

$x+{50}^{\circ }={180}^{\circ }$

$x={130}^{\circ }$

$x=y={130}^{\circ }$

According to the given figure, we can conclude that $x$ and $y$ form a pair of alternate interior angles parallel to the lines AB and CD.

Hence, we can conclude that $x={130}^{\circ },y={130}^{\circ }$ and.

4. In the given figure, if $\text{AB}||\text{CD},\text{CD}||\text{EF}$ and $\text{y}:\text{z}=3:7$, find $\text{x}$.

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Ans: According to the question given that,  and $y:z=3:7$.

We need to find the value of $x$ in the figure given below.

As we also know that the lines parallel to the same line are also parallel to each other.

We can determine that .

Assume that, $y=3a$ and $z=7a$.

We know that angles on same side of a transversal are supplementary.

So, $x+y={180}^{\circ }.$

$x=z$ (Alternate interior angles)

$z+y={180}^{\circ }$, or $7a+3a={180}^{\circ }$

$\Rightarrow 10a={180}^{\circ }$

$a={18}^{\circ }$.

$z=7a={126}^{\circ }$

$y=3a={54}^{\circ }$

Now, $x+{54}^{\circ }={180}^{\circ }$

$x={126}^{\circ }$

Hence, we can determine that $x={126}^{\circ }$.

5. In the given figure, if  and $\mathrm{\angle }PRD={127}^{\circ }$, find $\text{x}$ and $\text{y}$.

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Ans: According to the question given that,  and $\mathrm{\angle }PRD={127}^{\circ }$.

As we need to find the value of $x$ and $y$ in the figure.

$\mathrm{\angle }APQ=x={50}^{\circ }$. (Alternate interior angles)

$\mathrm{\angle }PRD=\mathrm{\angle }APR={127}^{\circ }$. (Alternate interior angles)

$\mathrm{\angle }APR=\mathrm{\angle }QPR+\mathrm{\angle }APQ$

${127}^{\circ }=y+{50}^{\circ }$

$\Rightarrow y={77}^{\circ }$

Hence, we can determine that $x={50}^{\circ }$ and $y={77}^{\circ }$.

6. In the given figure, sides QP and RQ of $\mathrm{\Delta }\text{PQR}$ are produced to points $\text{S}$ and $\text{T}$ respectively. If $\mathrm{\angle }\text{SPR}={135}^{\circ }$ and $\mathrm{\angle }\text{PQT}={110}^{\circ }$, find $\mathrm{\angle }\text{PRQ}$.

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Ans: According to the question given that, $\mathrm{\angle }SPR={135}^{\circ }$ and $\mathrm{\angle }PQT={110}^{\circ }$.

As we need to find the value of $\mathrm{\angle }PRQ$ in the figure given below.

According to the given figure, we can determine that $\mathrm{\angle }SPR$ and $\mathrm{\angle }RPQ$, and $\mathrm{\angle }SPR$ and $\mathrm{\angle }RPQ$ form a linear pair.

As we also know that the sum of angles of a linear pair is ${180}^{\circ }$.

$\mathrm{\angle }SPR+\mathrm{\angle }RPQ={180}^{\circ }$, and

$\mathrm{\angle }PQT+\mathrm{\angle }PQR={180}^{\circ }$

${135}^{\circ }+\mathrm{\angle }RPQ={180}^{\circ }$, and

${110}^{\circ }+\mathrm{\angle }PQR={180}^{\circ }$, or

$\mathrm{\angle }RPQ={45}^{\circ }$, and

$\mathrm{\angle }PQR={70}^{\circ }.$

According to the figure, we can determine that

$\mathrm{\angle }PQR+\mathrm{\angle }RPQ+\mathrm{\angle }PRQ={180}^{\circ }$. (Angle sum property)

$\Rightarrow {70}^{\circ }+{45}^{\circ }+\mathrm{\angle }PRQ={180}^{\circ }$

$\Rightarrow {115}^{\circ }+\mathrm{\angle }PRQ={180}^{\circ }$

$\Rightarrow {115}^{\circ }+\mathrm{\angle }PRQ={180}^{\circ }$

$\Rightarrow \mathrm{\angle }PRQ={65}^{\circ }$.

Hence, we can determine that $\mathrm{\angle }PRQ={65}^{\circ }$.

7. In the given figure, $\mathrm{\angle }\text{X}={62}^{\circ },\mathrm{\angle }\text{XYZ}={54}^{\circ }.$ If $\text{YO}$ and $\text{ZO}$ are the bisectors of $\mathrm{\angle }\text{XYZ}$ and $\mathrm{\angle }\text{XZY}$ respectively of $\mathrm{\Delta }\mathbf{XYZ}$, find $\mathrm{\angle }\text{OZY}$ and $\mathrm{\angle }\text{YOZ}$.

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Ans: According to the question given that, $\mathrm{\angle }X={62}^{\circ },\mathrm{\angle }XYZ={54}^{\circ }$ and YO and ZO are bisectors of $\mathrm{\angle }XYZ$ and $\mathrm{\angle }XZY$, respectively.

As we need to find the value of$\mathrm{\angle }OZY$ and $\mathrm{\angle }YOZ$ in the figure.

According to the given figure, we can determine that in $\mathrm{\Delta }XYZ$

$\mathrm{\angle }X+\mathrm{\angle }XYZ+\mathrm{\angle }XZY={180}^{\circ }$ (Angle sum property)

$\Rightarrow {62}^{\circ }+{54}^{\circ }+\mathrm{\angle }XZY={180}^{\circ }$

$\Rightarrow {116}^{\circ }+\mathrm{\angle }XZY={180}^{\circ }$

$\Rightarrow \mathrm{\angle }XZY={64}^{\circ }$.

According to the question given that, OY and OZ are the bisectors of $\mathrm{\angle }XYZ$ and $\mathrm{\angle }XZY$, respectively.

$\mathrm{\angle }OYZ=\mathrm{\angle }XYO={\displaystyle \frac{{54}^{\circ }}{2}}={27}^{\circ }$, and

$\mathrm{\angle }OZY=\mathrm{\angle }XZO={\displaystyle \frac{{64}^{\circ }}{2}}={32}^{\circ }$

According to the figure, we can determine that in $\mathrm{\Delta }OYZ$

$\mathrm{\angle }OYZ+\mathrm{\angle }OZY+\mathrm{\angle }YOZ={180}^{\circ }$. (Angle sum property)

${27}^{\circ }+{32}^{\circ }+\mathrm{\angle }YOZ={180}^{\circ }$

$\Rightarrow {59}^{\circ }+\mathrm{\angle }YOZ={180}^{\circ }$

$\Rightarrow \mathrm{\angle }YOZ={121}^{\circ }$.

Hence, we can determine that $\mathrm{\angle }YOZ={121}^{\circ }$ and $\mathrm{\angle }OZY={32}^{\circ }\text{. }$

8. In the given figure, if $\text{AB}||\text{DE},\mathrm{\angle }\text{BAC}={35}^{\circ }$ and $\mathrm{\angle }\text{CDE}={53}^{\circ }$, find $\mathrm{\angle }\text{DCE}$.

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Ans: According to the question given that,  and $\mathrm{\angle }CDE={53}^{\circ }$.

As we need to find the value of $\mathrm{\angle }DCE$ in the figure given below.

According to the given figure, we can determine that

$\mathrm{\angle }BAC=\mathrm{\angle }CED={35}^{\circ }$ (Alternate interior angles)

According to the figure, we can determine that in $\mathrm{\Delta }DCE$

$\mathrm{\angle }DCE+\mathrm{\angle }CED+\mathrm{\angle }CDE={180}^{\circ }$. (Angle sum property)

$\mathrm{\angle }DCE+{35}^{\circ }+{53}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DCE+{88}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DCE={92}^{\circ }$.

Hence, we can determine that $\mathrm{\angle }DCE={92}^{\circ }$.

9. In the given figure, if lines PQ and RS intersect at point $\text{T}$, such that $\mathrm{\angle }\text{PRT}={40}^{\circ }$, $\mathrm{\angle }\text{RPT}={95}^{\circ }$ and $\mathrm{\angle }\text{TSQ}={75}^{\circ }$, find $\mathrm{\angle }\text{SQT}$.

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Ans: According to the question given that, $\mathrm{\angle }PRT={40}^{\circ },\mathrm{\angle }RPT={95}^{\circ }$ and $\mathrm{\angle }TSQ={75}^{\circ }$.

As we need to find the value of $\mathrm{\angle }SQT$ in the figure.

According to the given figure, we can determine that in $\mathrm{\Delta }RTP$

$\mathrm{\angle }PRT+\mathrm{\angle }RTP+\mathrm{\angle }RPT={180}^{\circ }$ (Angle sum property)

${40}^{\circ }+\mathrm{\angle }RTP+{95}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }RTP+{135}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }RTP={45}^{\circ }$.

According to the figure, we can determine that

$\mathrm{\angle }RTP=\mathrm{\angle }STQ={45}^{\circ }.$ (Vertically opposite angles)

According to the figure, we can determine that in $\mathrm{\Delta }STQ$

$\mathrm{\angle }SQT+\mathrm{\angle }STQ+\mathrm{\angle }TSQ={180}^{\circ }$. (Angle sum property)

$\mathrm{\angle }SQT+{45}^{\circ }+{75}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }SQT+{120}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }SQT={60}^{\circ }$.

Hence, we can determine that $\mathrm{\angle }SQT={60}^{\circ }$.

10. In fig lines $\text{XY}$ and $\text{MN}$ intersect at O If $\mathrm{\angle }$ POY $={90}^{\circ }$ and $\text{ab}=2:3$ find $\text{c}$

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Ans: According to the given figure $\mathrm{\angle }\text{POY}={90}^{\circ }$

a: b: 2: 3

Assume that, $a=2x$ and $b=3x$

$\text{a}+\text{b}+\mathrm{\angle }\text{POY}={180}^{\circ }(\because \text{XOY}$ is a line $)$

$\Rightarrow$ $2\text{x}+3\text{x}+{90}^{\circ }={180}^{\circ }$

$\Rightarrow$ $5x={180}^{\circ }-{90}^{\circ }$

$\Rightarrow$ $5\text{x}={90}^{\circ }$

$\Rightarrow$ $x={\displaystyle \frac{{90}^{\circ }}{5}}={18}^{\circ }$

So, $a={36}^{\circ },b={54}^{\circ }$

MON is a line.

$\text{b}+\text{c}={180}^{\circ }$

$\Rightarrow$ ${54}^{\circ }+{\text{c}}^{\circ }={180}^{\circ }$

$\Rightarrow$ $c={180}^{\circ }-{54}^{\circ }={126}^{\circ }$

Hence, the value of $c={126}^{\circ }$.

11. In fig find the volume of $\text{x}$ and $\text{y}$ then Show that $\mathrm{AB}\Vert \mathrm{CD}$

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Ans: Accordign to the given figure, ${50}^{\circ }+x={180}^{\circ }$

(by linear pair)

$x={180}^{\circ }-{50}^{\circ }$

So, $x={130}^{\circ }$

$y={130}^{\circ }$ (Because vertically opposite angles are equal)

$\text{x}=\text{y}$ as they are corresponding angles. 

So, AB || CD

Hence proved.

12. What value of $\mathbf{y}$ would make AOB a line if $\mathrm{\angle }\text{AOC}=4\text{y}$ and $\mathrm{\angle }\text{BOC}=6\text{y}+{30}^{\circ }$

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Ans: According to the question given that, $\mathrm{\angle }\text{AOC}=4\text{y}$ and $\mathrm{\angle }\text{BOC}=6\text{y}+{30}^{\circ }$

$\mathrm{\angle }\text{AOC}+\mathrm{\angle }\text{BOC}={180}^{\circ }$     (By linear pair)

$4y+6y+{30}^{\circ }={180}^{\circ }$

$\Rightarrow$ $10y={180}^{\circ }-{30}^{\circ }$

$\Rightarrow$  $10y={150}^{\circ }$

$\Rightarrow$  $y={15}^{\circ }$

13. In fig $\text{POQ}$ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $\mathrm{\angle }\text{ROS}={\displaystyle \frac{1}{2}}(\mathrm{\angle }\text{QOS}-\mathrm{\angle }\text{POS})$

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Ans: According to the question,

$R.H.S={\displaystyle \frac{1}{2}}(\mathrm{\angle }QOS-\mathrm{\angle }POS)$

$={\displaystyle \frac{1}{2}}(\mathrm{\angle }\text{ROS}+\mathrm{\angle }\text{QOR}-\mathrm{\angle }\text{POS})$

$={\displaystyle \frac{1}{2}}\left(\mathrm{\angle }\text{ROS}+{90}^{\circ }-\mathrm{\angle }\text{POS}\right)\dots \dots ..$ (1)

Because,$\mathrm{\angle }\text{POS}+\mathrm{\angle }\text{ROS}={90}^{\circ }$

So, by equation 1

$={\displaystyle \frac{1}{2}}(ROS+\mathrm{\angle }POS+\mathrm{\angle }ROS-\mathrm{\angle }POS)$ (by equation 1)

$={\displaystyle \frac{1}{2}}\times 2\mathrm{\angle }\text{ROS}=\mathrm{\angle }\text{ROS}$

= L.H.S

Hence proved.

14. In fig lines l1and l2 intersected at O , if $\text{x}={45}^{\circ }$ find $\text{x},\text{y}$ and $\text{u}$

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Ans: According to the question given that,

$x={45}^{\circ }$

So, $\text{z}={45}^{\circ }$ (Because vertically opposite angles are equal)

$\text{x}+\text{y}={180}^{\circ }$

${45}^{\circ }+y={180}^{\circ }$ (By linear pair)

$y={180}^{\circ }-45$

$y={135}^{\circ }$

$\text{y}=\text{u}$

Hence, the value of $\text{u}={135}^{\circ }$ (Vertically opposite angles)

15. The exterior angle of a triangle is ${110}^{\circ }$ and one of the interior opposite angle is ${35}^{\circ }$. Find the other two angles of the triangle.

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Ans: As we all know that the exterior angle of a triangle is equal to the sum of interior opposite angles.

So, $\mathrm{\angle }\text{ACD}=\mathrm{\angle }\text{A}+\mathrm{\angle }\text{B}$

${110}^{\circ }=\mathrm{\angle }\text{A}+{35}^{\circ }$

$\Rightarrow$ $\mathrm{\angle }\text{A}={110}^{\circ }-{35}^{\circ }$

$\Rightarrow$ $\mathrm{\angle }\text{A}={75}^{\circ }$

$\Rightarrow$ $\mathrm{\angle }\text{C}=180-(\mathrm{\angle }\text{A}+\mathrm{\angle }\text{B})$

$\Rightarrow$ $\mathrm{\angle }\text{C}=180-\left({75}^{\circ }+{35}^{\circ }\right)$

$\mathrm{\angle }\text{C}={70}^{\circ }$

16. Of the three angles of a triangle, one is twice the smallest and another is three times the smallest. Find the angles.

Ans: Assume that the smallest angle be $x^{\circ }$

Then the other two angles are $2x^{\circ }$ and $3x^{\circ }$

$x^{\circ }+2x^{\circ }+3x^{\circ }={180}^{\circ }$ As we know that the sum of three angle of a triangle is $\text{Missing or unrecognized delimiter for right}$

$6x^{\circ }={180}^{\circ }$

$\text{x}={\displaystyle \frac{180}{6}}$

$={30}^{\circ }$

Hence, angles are ${30}^{\circ },{60}^{\circ }$ and ${90}^{\circ }$.

17. Prove that if one angle of a triangle is equal to the sum of other two angles, the triangle is right angled.

Ans: According to the question given that in $\mathrm{\Delta }ABC,\mathrm{\angle }\text{B}=\mathrm{\angle }\text{A}+\mathrm{\angle }\text{C}$

To prove: $\mathrm{\Delta }ABC$ is right angled.

Proof: $\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }\dots ..$ (1) (As we know that the sum of three angles of a $\mathrm{\Delta }\text{ABC}$ is ${180}^{\circ })$

$\mathrm{\angle }A+\mathrm{\angle }C=\mathrm{\angle }B\dots ..$ (2)

From equations (1) and (2),

$\mathrm{\angle }B+\mathrm{\angle }B={180}^{\circ }$

$2\mathrm{\angle }\text{B}={180}^{\circ }$

$\mathrm{\angle }\text{B}={90}^{\circ }$

18. In fig. sides QP and RQ of $\mathrm{\Delta }PQR$ are produced to points $\text{S}$ and $\text{T}$ respectively. If $\mathrm{\angle }SPR={135}^{\circ }$ and $\mathrm{\angle }\text{PQT}={110}^{\circ }$, find $\mathrm{\angle }\text{PRQ}$.

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Ans: According to the given figure,

$\mathrm{\angle }PQT+\mathrm{\angle }PQR={180}^{\circ }$

${110}^{\circ }+\mathrm{\angle }PQR={180}^{\circ }$

$\mathrm{\angle }PQR={180}^{\circ }-{110}^{\circ }$

$\mathrm{\angle }\text{PQR}={70}^{\circ }$

Also, $\mathrm{\angle }\text{SPR}=\mathrm{\angle }\text{PQR}+\mathrm{\angle }\text{PRQ}$ (According to the Interior angle theorem)

${135}^{\circ }={70}^{\circ }+\mathrm{\angle }PRQ$

$\mathrm{\angle }\text{PRQ}={135}^{\circ }-{70}^{\circ }$

Hence, the value of $\mathrm{\angle }\text{PRQ}={65}^{\circ }$.

19. In fig the bisector of $\mathrm{\angle }ABC$ and $\mathrm{\angle }\text{BCA}$ intersect each other at point O prove that $\mathrm{\angle }BOC={90}^{\circ }+{\displaystyle \frac{1}{2}}\mathrm{\angle }A$

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Ans: According to the question given that in $△ABC$ such that the bisectors of $\mathrm{\angle }ABC$ and $\mathrm{\angle }\text{BCA}$ meet at a point O.

To Prove $\mathrm{\angle }BOC={90}^{\circ }+{\displaystyle \frac{1}{2}}\mathrm{\angle }A$

Proof: In $△BOC$

$\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }BOC={180}^{\circ }$ (1)

In $△ABC$

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

$\mathrm{\angle }A+2\mathrm{\angle }1+2\mathrm{\angle }2={180}^{\circ }$

(BO and CO bisects $\mathrm{\angle }B$ and $\mathrm{\angle }\text{C}$ )

$\Rightarrow {\displaystyle \frac{\mathrm{\angle }A}{2}}+\mathrm{\angle }1+\mathrm{\angle }2={90}^{\circ }$

$\mathrm{\angle }1+\mathrm{\angle }2={90}^{\circ }-{\displaystyle \frac{\mathrm{\angle }A}{2}}$

(Divide forth side by 2)

$\mathrm{\angle }1+\mathrm{\angle }2={90}^{\circ }-{\displaystyle \frac{\mathrm{\angle }A}{2}}$ in (i)

Substituting, ${90}^{\circ }-{\displaystyle \frac{\mathrm{\angle }A}{2}}+\mathrm{\angle }BOC={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BOC={90}^{\circ }+{\displaystyle \frac{\mathrm{\angle }A}{2}}$

Hence proved.

20. In the given figure $\mathrm{\angle }POR$ and $\mathrm{\angle }QOR$ form a linear pair if $\mathbf{a}-\mathbf{b}={80}^{\circ }$. Find the

value of 'a' and 'b'.

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Ans: $a+b={180}^{\circ }\to (1)$ (By line as pair)

$a-b={80}^0\to (2)$

$2\text{a}={260}^{\circ }$ (Adding equations (1) and (2))

$\text{a}={130}^{\circ }$

Put $\text{a}={130}^{\circ }$ in equation (1)

${130}^{\circ }+b={180}^{\circ }$

$\text{b}={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Hence the value of $\text{a}={130}^{\circ }$ and $\text{b}={50}^{\circ }$.

21. If ray $\text{OC}$ stands on a line $\text{AB}$ such that $\mathrm{\angle }AOC=\mathrm{\angle }BOC$, then show that $\mathrm{\angle }AOC={90}^{\circ }$

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Ans: According to the question given that,

$\mathrm{\angle }AOC=\mathrm{\angle }BOC$ 

$\mathrm{\angle }\text{AOC}+\mathrm{\angle }\text{BOC}={180}^{\circ }$   (By lines pair)

$\mathrm{\angle }\text{AOC}+\mathrm{\angle }\text{AOC}={180}^{\circ }$

$2\mathrm{\angle }\text{AOC}={180}^{\circ }$

$\mathrm{\angle }\text{AOC}={90}^{\circ }=\mathrm{\angle }B\text{OC}$

22. In the given figure show that $\mathrm{AB}\Vert \mathrm{EF}$

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Ans: $\mathrm{\angle }\text{BCD}=\mathrm{\angle }\text{BCE}+\mathrm{\angle }\text{ECD}$

$={36}^{\circ }+{30}^{\circ }={66}^{\circ }=\mathrm{\angle }ABC$

So,  (Alternate interior angles are equal)

Again, $\mathrm{\angle }\text{ECD}={30}^{\circ }$ and $\mathrm{\angle }\text{FEC}={150}^{\circ }$

So, $\mathrm{\angle }\text{ECD}+\mathrm{\angle }\text{FEC}={30}^{\circ }+{150}^{\circ }={180}^{\circ }$

Therefore,  (We know that the sum of consecutive interior angle is ${180}^{\circ })$

$AB\Vert CD$ and $\mathrm{CD}\Vert \mathrm{EF}$

Then $\mathrm{AB}\Vert \mathrm{EF}$

Hence proved.

23. In figure if  and $\mathrm{\angle }PRD={127}^{\circ }$ Find $\text{x}$ and $\text{y}$.

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Ans: $AB\Vert CD$ and PQ is a transversal

$\mathrm{\angle }\text{APQ}=\mathrm{\angle }\text{PQD}$ (Pair of alternate angles)

${50}^{\circ }=\text{x}$

Also  and $\text{PR}$ is a transversal

$\mathrm{\angle }\text{APR}=\mathrm{\angle }\text{PRD}$

${50}^{\circ }+Y={127}^{\circ }$

$\text{Y}={127}^{\circ }-{50}^{\circ }={77}^{\circ }$

Hence the value of $\text{x}={50}^{\circ }$ and $\text{Y}={77}^{\circ }$.

24. Prove that if two lines intersect each other then vertically opposite angler are equal.

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Ans: According to the given figure: AB and CD are two lines intersect each other at $O$.

To prove: (i) $\mathrm{\angle }1=\mathrm{\angle }2$ and (ii) $\mathrm{\angle }3=\mathrm{\angle }4$

Proof:

$\mathrm{\angle }1+\mathrm{\angle }4={180}^{\circ }\to (i)$ (By linear pair)

$\mathrm{\angle }4+\mathrm{\angle }2={180}^{\circ }\to (ii)$

$\mathrm{\angle }1+\mathrm{\angle }4=\mathrm{\angle }4+\mathrm{\angle }2$ (By equations (i) and (ii))

$\mathrm{\angle }1=\mathrm{\angle }2$

Similarly,

$\mathrm{\angle }3=\mathrm{\angle }4$

Hence proved.

25. The measure of an angle is twice the measure of the supplementary angle. Find measure of angles.

Ans: Assume that the measure be $x^{\circ }$.

Then its supplement is ${180}^{\circ }-x^{\circ }$.

According to question

$x^{\circ }=2\left({180}^{\circ }-x^{\circ }\right)$

$\Rightarrow$ $x^{\circ }={360}^{\circ }-2x^{\circ }$

$\Rightarrow$  $3x={360}^{\circ }$

$\Rightarrow$  $x={120}^{\circ }$

The measure of the angles are ${120}^{\circ }$ and ${60}^{\circ }$.

26. In fig $\mathrm{\angle }PQR=\mathrm{\angle }PRQ$. Then prove that $\mathrm{\angle }PQS=\mathrm{\angle }PRT$.

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Ans: $\mathrm{\angle }PQS+\mathrm{\angle }PQR=\mathrm{\angle }PRQ+\mathrm{\angle }PRT$ (By linear pair)

But,

$\mathrm{\angle }PQR=\mathrm{\angle }PRQ$ (Accordign to the question)

So, $\mathrm{\angle }PQS=\mathrm{\angle }PRT$

Hence proved.

27. In the given fig $\mathrm{\angle }\text{AOC}=\mathrm{\angle }\text{ACO}$ and $\mathrm{\angle }\text{BOD}=\mathrm{\angle }\text{BDO}$ prove that AC || DB.

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Ans: According to the question given that,

$\mathrm{\angle }AOC=\mathrm{\angle }ACO$ and $\mathrm{\angle }BOD=\mathrm{\angle }BDO$

But,

$\mathrm{\angle }AOC=\mathrm{\angle }BOD$ (Vertically opposite angles)

$\mathrm{\angle }AOC=\mathrm{\angle }BOD$ and $\mathrm{\angle }BOD=\mathrm{\angle }BDO$

$\Rightarrow \mathrm{\angle }ACO=\mathrm{\angle }BDO$

So,  (By alternate interior angle property)

Hence AC || DB  proved.

28. In figure if lines $\text{PQ}$ and $\text{RS}$ intersect at point $\text{T}$. Such that $\mathrm{\angle }PRT={40}^{\circ }$, $\mathrm{\angle }RPT={95}^{\circ }$ and $\mathrm{\angle }TSQ={75}^{\circ }$, find $\mathrm{\angle }SQT$.

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Ans: According to the $\mathrm{\Delta }$ PRT

$\mathrm{\angle }\text{P}+\mathrm{\angle }\text{R}+\mathrm{\angle }1={180}^{\circ }$ (By angle sum property)

${95}^{\circ }+{40}^{\circ }+\mathrm{\angle }1={180}^{\circ }$

$\mathrm{\angle }1={180}^{\circ }-{135}^{\circ }$

$\mathrm{\angle }1={45}^{\circ }$

$\mathrm{\angle }1=\mathrm{\angle }2$ (Vertically opposite angle)

$\mathrm{\angle }2=\mathrm{\angle }{45}^{\circ }$

According to the $\mathrm{\Delta }\text{TQS}\mathrm{\angle }2+\mathrm{\angle }\text{Q}+\mathrm{\angle }\text{S}={180}^{\circ }$

${45}^{\circ }+\mathrm{\angle }Q+{75}^{\circ }={180}^{\circ }$

$\mathrm{\angle }\text{Q}+{120}^{\circ }={180}^{\circ }$

$\mathrm{\angle }\text{Q}={180}^{\circ }-{120}^{\circ }$

$\mathrm{\angle }\text{Q}={60}^{\circ }$

Hence, the value of $\mathrm{\angle }\text{SQT}={60}^{\circ }$.

29. In figure, if $QT\mathrm{\perp }PR,\mathrm{\angle }TQR={40}^{\circ }$ and $\mathrm{\angle }SPR={50}^{\circ }$ find $x$ and $y$.

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Ans: According to the $\mathrm{\Delta }\text{TQR}$

${90}^{\circ }+{40}^{\circ }+x={180}^{\circ }$ (Angle sum property of triangle)

So,$x={50}^{\circ }$

Now, $y=\mathrm{\angle }\text{SPR}+x$

So, $y={30}^{\circ }+{50}^{\circ }={80}^{\circ }$.

Hence, the value of $x={50}^{\circ }$ and $y={80}^{\circ }$.

30. In figure sides QP and RQ of $\mathrm{\Delta }PQR$ are produced to points $\text{S}$ and $\text{T}$ respectively if $\mathrm{\angle }SPR={135}^{\circ }$ and $\mathrm{\angle }PQT={110}^{\circ }$, find $\mathrm{\angle }PRQ$.

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Ans: According to the given figure,

${110}^{\circ }+\mathrm{\angle }2={180}^{\circ }$ (By linear pair)

$\mathrm{\angle }2={180}^{\circ }-{110}^{\circ }$

$\mathrm{\angle }2={70}^{\circ }$

$\mathrm{\angle }1+{135}^{\circ }={180}^{\circ }$

$\mathrm{\angle }1={180}^{\circ }-{135}^{\circ }$

$\mathrm{\angle }1={45}^{\circ }$

$\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }\text{R}={180}^{\circ }$ (By angle sum property)

${45}^{\circ }+{70}^{\circ }+\mathrm{\angle }R={180}^{\circ }$

$\mathrm{\angle }\text{R}={180}^{\circ }-{115}^{\circ }$

$\mathrm{\angle }\text{R}={65}^{\circ }$

Hence, the value of $\mathrm{\angle }\text{PRQ}={65}^{\circ }$.

31. In figure lines $\text{PQ}$ and RS intersect each other at point O. If $\mathrm{\angle }POR:\mathrm{\angle }ROQ=5:7$. Find all the angles.

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Ans: $\mathrm{\angle }POR+\mathrm{\angle }ROQ={180}^{\circ }$ (Linear pair of angle)

But, $\mathrm{\angle }\text{POR}:\mathrm{\angle }\text{ROQ}=5:7$ (According to the question)

So, $\mathrm{\angle }\text{POR}={\displaystyle \frac{5}{12}}\times {180}^{\circ }={75}^{\circ }$

Similarly, $\mathrm{\angle }\text{ROQ}={\displaystyle \frac{7}{12}}\times {180}^{\circ }={105}^{\circ }$