# Grade 9 – Math – Lines and Angles – Important Questions and Answers

Question 1: In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that,

$\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})$

$\begin{array}{ll} \angle \mathrm{ROS}=\angle \mathrm{ROP}-\angle \mathrm{POS} & \ldots(1) \\ \angle \mathrm{ROS}=\angle \mathrm{QOS}-\angle \mathrm{QOR} & \ldots(2)\end{array}$

$\angle \mathrm{ROS}+\angle \mathrm{ROS}=\angle \mathrm{QOS}-\angle \mathrm{QOR}+\angle \mathrm{ROP}-\angle \mathrm{POS}$
$\Rightarrow 2 \angle \mathrm{ROS}=\angle \mathrm{QOS}-\angle \mathrm{POS}\hspace{1cm}(\because \angle \mathrm{QOR}=\angle \mathrm{ROP}=90^{\circ})$
$\Rightarrow \angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS}) \quad$ Proved.

Question 2: It is given that $\angle X Y Z=64^{\circ}$ and $X Y$ is produced to a point P. Draw a figure from the given information. If ray $Y Q$ bisects $\angle$ ZYP, find $\angle X Y Q$ and reflex $\angle Q Y P$ .

Answer. From figure,$\angle \text{XYZ}=64^{\circ} \quad[\text { Given }]$

Question 3. Let $O A, O B, O C$ and $O D$ are rays in the anti-clockwise direction such that $\angle A O B=\angle C O D=100^{\circ}, \angle B O C=82^{\circ}$ and $\angle A O D=78^{\circ}$ Is it true to say that AOC and BOD are lines?

Answer. $\angle \mathrm{AOB}+\angle \mathrm{BOC}=100^{\circ}+82^{\circ}=182^{\circ} \neq 180^{\circ}$
Hence, AOC is not a line.
$\angle \mathrm{BOC}+\angle \mathrm{COD}=82^{\circ}+100^{\circ}=182^{\circ} \neq 180^{\circ}$
Hence, BOD is not a line.

Question 4: In the given figure, $\angle{AOC}$ and $\angle{BOC}$ form a linear pair. If $a – b = 80°$, find the values of $a$ and $b$.

We have, $a – b = 80°$ [Given]
Also, $a + b = 180°$ [Linear pair]
$\Rightarrow 2a = 260°$
$\Rightarrow a = 130°$ and
$b = 180° – 130° = 50°$.
So, $a$ and $b$ are 130° and 50° respectively.

Question 5:. In the given figure, if $\angle C O D=90^{\circ},$ then find$\angle A O C$ and $\angle B O D .$

Answer:. since $\mathrm{AOB}$ is a line, therefore, $\angle \mathrm{AOB}+\angle \mathrm{COD}+\angle \mathrm{BOD}=180^{\circ}$
$\Rightarrow 2 x+90^{\circ}+x=180^{\circ} \Rightarrow 3 x=90^{\circ} \Rightarrow x=30^{\circ}$
$\therefore 2 x=2 \times 30^{\circ}=60^{\circ}$
$\therefore \angle \mathrm{AOC}=60^{\circ}$ and $\angle \mathrm{BOD}=30^{\circ}$

Question 5:. In the given figure, ray $O E$ bisects $\angle A O B$ and $O F$ is a ray opposite to OE. Show that $\angle F O B=\angle F O A$

Since ray OE bisects $\angle \mathrm{AOB}$
$\therefore\angle \mathrm{EOB}+\angle \mathrm{EOA}\ldots$(i)
Now, ray OB stands on the line EF.
$\therefore \angle \mathrm{EOB}+\angle \mathrm{FOB}=180^{\circ} \ldots$ (ii) $[$ Linear pair $]$
Again, ray OA stands on the line EF.
$\therefore \angle \mathrm{EOA}+\angle \mathrm{FOA}=180^{\circ}\ldots$(iii)
From (ii) and (iii), we get
$\angle \mathrm{EOB}+\angle \mathrm{FOB}=\angle \mathrm{EOA}+\angle \mathrm{FOA}$
$\Rightarrow \angle \mathrm{EOA}+\angle \mathrm{FOB}=\angle \mathrm{EOA}+\angle \mathrm{FOA}\hspace{1cm}[\because \angle \mathrm{EOB}=\angle \mathrm{EOA} \text { from }\text{(ii)}]$
$\Rightarrow \angle \mathrm{FOB}=\angle \mathrm{FOA}$

Question 6: If ray $O C$ stands on line $A B$ such that $\angle A O C=\angle C O B,$ then show that $\angle A O C=90^{\circ}$

Answer. since ray $\mathrm{OC}$ stands on line $\mathrm{AB}$. Therefore, $\angle \mathrm{AOC}+\angle \mathrm{COB}=180^{\circ} \quad[$ Linear pair $]$
But $\angle \mathrm{AOC}=\angle \mathrm{COB}\hspace{0.5cm}$[Given]
$\therefore \angle \mathrm{AOC}+\angle \mathrm{AOC}=180^{\circ}$
$\Rightarrow 2 \angle \mathrm{AOC}=180^{\circ}$
$\Rightarrow \angle \mathrm{AOC}=90^{\circ}$

Question 7: In the figure, ray OX bisects $\angle \mathrm{AOC}$ and ray OY bisects $\angle B O C$. If $\angle \mathrm{XOY}=90^{\circ}$, show that A, $O, B$ are collinear.

Answer. We have, $\angle \mathrm{AOC}=2 \angle \mathrm{XOC}\hspace{1cm}\ldots$(i)
$[\because$ ray $\mathrm{OX}$ bisects $\angle \mathrm{AOC}]$
$\mathrm{And}, \angle \mathrm{BOC}=2 \angle \mathrm{YOC}\hspace{1cm}\ldots$(ii)
$[\because \text { ray OY bisects } \angle \mathrm{BOC}]$
Adding, (i) and (ii) we get
$\begin{array}{l} \angle \mathrm{AOC}+\angle \mathrm{BOC}=2 \angle \mathrm{XOC}+2 \angle \mathrm{YOC}\\ =2(\angle \mathrm{XOC}+\angle \mathrm{YOC}) \\ =2 \angle \mathrm{XOY}\\=2 \times 90^{\circ} &\left(\because \angle \mathrm{XOY}=90^{\circ}\right) \\ = 180^{\circ} \end{array}$
Therefore, $\angle \mathrm{AOC}$ and $\angle \mathrm{BOC}$ form a linear pair. Consequently OA and OB are two opposite rays. Hence, $\mathrm{A}, \mathrm{O}, \mathrm{B}$ are collinear.

Question 8: If two lines intersect, prove that the vertically opposite angles are equal.

Given : Two lines AB and CD intersect at O
To prove :
(i) $\angle \mathrm{AOC}=\angle \mathrm{BOD}$
(ii) $\angle \mathrm{AOD}=\angle \mathrm{BOC}$

Proof : since ray OC stands on the line AB, we have
$\angle \mathrm{AOC}+\angle \mathrm{COB}=180^{\circ} \quad \ldots$ (i) $[$ Linear pair $]$
since, ray OA stands on the line $\mathrm{CD}$, we have
$\angle \mathrm{AOC}+\angle \mathrm{AOD}=180^{\circ}\hspace{1cm} \ldots$(ii) [Linear pair]
From (i) and (ii), we have $\angle \mathrm{AOC}+\angle \mathrm{COB}=\angle \mathrm{AOC}+\angle \mathrm{AOD}$
$\Rightarrow \angle \mathrm{COB}=\angle \mathrm{AOD}$
Similarly, $\angle \mathrm{AOC}=\angle \mathrm{BOD}$.

Question 9: In the given figure, two straight lines PQ and RS intersect each other at O. If $\angle{POT} = 75°$. Find the values of b and c.

Answer. since $O R$ and $O S$ are in the same line.
Therefore, $\angle \mathrm{ROP}+\angle \mathrm{POT}+\angle \mathrm{TOS}=180^{\circ}$
$\Rightarrow 4 b+75^{\circ}+b=180^{\circ}$
$\Rightarrow 5 b+75^{\circ}=180^{\circ}$
$\Rightarrow 5 b=105^{\circ}$
$\Rightarrow b=21$
since, $P Q$ and $R S$ intersect at $O$. Therefore, $\angle \mathrm{QOS}=\angle \mathrm{POR} \quad$ [Vertically opp. angles]
$\Rightarrow a=4 a$
$\Rightarrow a=4 \times 21=84 \quad[\therefore b=21]$
Now, OR and OS are in the same line. Therefore, $\angle \mathrm{ROQ}+\angle \mathrm{QOS}=180^{\circ}$
$\Rightarrow 2 c+a=180^{\circ}$
$\Rightarrow 2 c+84=180^{\circ}$
$\Rightarrow 2 c=96$
$[\because a=84]$
$\Rightarrow c=48$
Hence, $b=21$ and $c=48$

Question 10: In the given figure, OP bisects $\angle \mathrm{BOC}$ and $O Q$ bisects $\angle \mathrm{A} O C$. Show that $\angle P O Q=90^{\circ}$

Answer: since OP bisects $\angle \mathrm{BOC}$
$\therefore \angle \mathrm{BOC}=2 \angle \mathrm{POC}\hspace{1cm}\ldots$(i)
Again, $\mathrm{OQ}$ bisects $\angle \mathrm{AOC}$
$\therefore \angle \mathrm{AOC}=2 \angle \mathrm{QOC}$
since ray $\mathrm{OC}$ stands on line $\mathrm{AB}$.
$\therefore \angle \mathrm{AOC}+\angle \mathrm{BOC}=180^{\circ}\hspace{1cm}\ldots$(i)
$\Rightarrow 2 \angle \mathrm{Q} \mathrm{OC}+2 \angle \mathrm{POC}=180^{\circ}[\mathrm{Using}(\mathrm{i})$ and $(\mathrm{ii})]$
$\Rightarrow 2(\angle \mathrm{POC}+\angle \mathrm{QOC})=180^{\circ}$
$\Rightarrow \angle \mathrm{POC}+\angle \mathrm{QOC}=90^{\circ}$
$\Rightarrow \angle P O Q=90^{\circ}, \quad$ Proved.

Question 11: In the figure, find the values of x and y and then show that AB || CD.

In the given figure, a transversal intersects two lines AB and CD such that x + 50° = 180° [Linear pair axiom]
$\Rightarrow \text{x} = 180° – 50° = 130°$
y = 130° [Vertically opposite angles]
Therefore, x = y = 130° [Alternate angles]
∴ AB || CD [Converse of alternate angles axiom]
Proved.

Question 12:. In the figure, if ${A B}|| {C D}, C D || EF$and $y: z=3: 7,$ find $x .$

Answer: In the given figure, $\mathrm{AB}||\mathrm{CD}, \mathrm{CD}|| \mathrm{EF}$ and $y: z=3: 7$
Let $y=3 a$ and $z=7 a$
$\angle \mathrm{DHI}=y \quad$ [Vertically opposite angles $]$
$\angle \mathrm{DHI}+\angle \mathrm{FIH}=180^{\circ}[$ Interior angles on the same side of the transversal]
\begin{aligned}{l} \Rightarrow \quad y+z=180^{\circ} & \\ \Rightarrow 3 a+7 a=180^{\circ} \\ \Rightarrow \quad 10 a=180^{\circ} \Rightarrow a=18^{\circ} \end{aligned}
Also, $\quad x+y=180^{\circ}$
$\Rightarrow \quad x+54^{\circ}=180^{\circ}$
$\therefore \quad x=180^{\circ}-54^{\circ}=126^{\circ}$