NCERT Solutions for Exercise 6.1 Class 9 Maths Chapter 6 - Lines and Angles

Q1. In Fig. 6.13, lines AB and CD intersect at O. If $\mathrm{\angle }$ AOC + $\mathrm{\angle }$ BOE = 70° and $\mathrm{\angle }$ BOD = 40°, find $\mathrm{\angle }$ BOE and reflex $\mathrm{\angle }$ COE.

Solution:
Given that,
AB is a straight line. Lines AB and CD intersect at O. $\mathrm{\angle }AOC+\mathrm{\angle }BOE={70}^0$ and $\mathrm{\angle }$ BOD = ${40}^0$
Since AB is a straight line

$\therefore$ $\mathrm{\angle }$ AOC + $\mathrm{\angle }$ COE + $\mathrm{\angle }$ EOB = ${180}^0$

$\Rightarrow \mathrm{\angle }COE={180}^0-{70}^0={110}^0$ [since $\mathrm{\angle }AOC+\mathrm{\angle }BOE={70}^0$ ]

So, reflex $\mathrm{\angle }$ COE = ${360}^0-{110}^0={250}^0$

It is given that AB and CD intersect at O

Therefore, $\mathrm{\angle }$ AOC = $\mathrm{\angle }$ BOD [vertically opposite angle]

$\Rightarrow \mathrm{\angle }COA={40}^0$ [ GIven $\mathrm{\angle }$ BOD = ${40}^0$ ]

Also, $\mathrm{\angle }AOC+\mathrm{\angle }BOE={70}^0$

So, $\mathrm{\angle }$ BOE = ${30}^0$

Q2. In Fig. 6.14, lines XY and MN intersect at O. If $\mathrm{\angle }POY={90}^o$ and a : b = 2 : 3, find c.

Solution:
Given that,
Line XY and MN intersect at O and $\mathrm{\angle }$ POY = ${90}^0$ also $a:b=2:3\Rightarrow b=\frac{3a}{2}$ ..............(i)

Since XY is a straight line

Therefore, $$\begin{gathered} a+b+\mathrm{\angle }POY={180}^0 \\ a+b={180}^0-{90}^0={90}^0 \end{gathered}$$ ...........(ii)

Thus, from eq (i) and eq (ii), we get

$\Rightarrow \frac{3a}{2}+a={90}^0$

$\Rightarrow a={36}^0$

So, $b={54}^0$

Since $\mathrm{\angle }$ MOY = $\mathrm{\angle }$ c [vertically opposite angles]

$\mathrm{\angle }$ a + $\mathrm{\angle }$ POY = c

${126}^0=c$

Q3. In Fig. 6.15, $\mathrm{\angle }$ PQR = $\mathrm{\angle }$ PRQ, then prove that $\mathrm{\angle }$ PQS = $\mathrm{\angle }$ PRT.

Solution:
Given that,
ABC is a triangle such that $\mathrm{\angle }$ PQR = $\mathrm{\angle }$ PRQ and ST is a straight line.

Now, $\mathrm{\angle }$ PQR + $\mathrm{\angle }$ PQS = ${180}^0$ {Linear pair}............(i)

Similarly, $\mathrm{\angle }$ PRQ + $\mathrm{\angle }$ PRT = ${180}^0$ ..................(ii)

equating the eq (i) and eq (ii), we get

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

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

Hence proved.

Q4 In Fig. 6.16, if x + y = w + z , then prove that AOB is a line.

Solution:
Given that,

$x+y=z+w$ ..............(i)

It is known that, the sum of all the angles at a point = ${360}^0$

$\therefore$ $x+y+z+w={360}^0$ ..............(ii)

From eq (i) and eq (ii), we get

$$\begin{gathered} 2(x+y)={360}^0 \\ x+y={180}^0 \end{gathered}$$

Hence proved AOB is a line.

Solution:
Given that,
POQ is a line, OR $\perp$ PQ and $\mathrm{\angle }$ ROQ is a right angle.

Now, $\mathrm{\angle }$ POS + $\mathrm{\angle }$ ROS + $\mathrm{\angle }$ ROQ = ${180}^0$ [since POQ is a straight line]

$$\begin{gathered} \Rightarrow \mathrm{\angle }POS+\mathrm{\angle }ROS={90}^0 \\ \Rightarrow \mathrm{\angle }ROS={90}^0-\mathrm{\angle }POS \end{gathered}$$ .............(i)

and, $\mathrm{\angle }$ ROS + $\mathrm{\angle }$ ROQ = $\mathrm{\angle }$ QOS

$\mathrm{\angle }ROS=\mathrm{\angle }QOS-{90}^0$ ..............(ii)

Add the eq (i ) and eq (ii), we get

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

Hence proved.

Solution:

Given that,

$\mathrm{\angle }$ XYZ = ${64}^0$ and XY produced to point P and Ray YQ bisects $\mathrm{\angle }$ ZYP $\Rightarrow \mathrm{\angle }QYP=\mathrm{\angle }ZYQ$
Now, XYP is a straight line

So, $\mathrm{\angle }$ XYZ + $\mathrm{\angle }$ ZYQ + $\mathrm{\angle }$ QYP = ${180}^0$

$\Rightarrow 2(\mathrm{\angle }QYP)={180}^0-{64}^0={116}^0$

$\Rightarrow (\mathrm{\angle }QYP)={58}^0$

Thus reflex of $\mathrm{\angle }$ QYP = ${360}^0-{58}^0={302}^0$

Since $\mathrm{\angle }$ XYQ = $\mathrm{\angle }$ XYZ + $\mathrm{\angle }$ ZYQ [ $\because$ $\mathrm{\angle }QYP=\mathrm{\angle }ZYQ$

$\mathrm{\angle }XYQ$ = ${64}^0+{58}^0={122}^0$