Exercise 6.1 (Revised) - Chapter 6 - Lines & Angles - Ncert Solutions class 9 - Maths

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NCERT Solutions for Class 9 Maths Chapter 6: Lines & Angles | Free PDF Download

Ex 6.1 Question 1.

In Fig. 6.13, lines $A B$ and $C D$ intersect at $O$. If $\angle A O C+\angle B O E=70^{\circ}$ and $\angle B O D=40^{\circ}$, find $\angle \mathrm{BOE}$ and reflex $\angle \mathrm{COE}$.

Answer.

We are given that $\angle A O C+\angle B O E=70^{\circ}$ and $\angle B O D=40^{\circ}$.

We need to find $\angle B O E$ and reflex $\angle C O E$.

From the given figure, we can conclude that $\angle C O B$ and $\angle C O E$ form a linear pair.

We know that sum of the angles of a linear pair is $180^{\circ}$.
$
\begin{aligned}
& \therefore \angle C O B+\angle C O E=180^{\circ} \\
& \because \angle C O B=\angle A O C+\angle B O E, \text { or } \\
& \therefore \angle A O C+\angle B O E+\angle C O E=180^{\circ} \\
& \Rightarrow 70^{\circ}+\angle C O E=180^{\circ} \\
& \Rightarrow \angle C O E=180^{\circ}-70^{\circ} \\
& =110^{\circ} . \\
& \text { Reflex } \angle C O E=360^{\circ}-\angle C O E \\
& =360^{\circ}-110^{\circ} \\
& =250^{\circ} .
\end{aligned}
$
$\angle A O C=\angle B O D$ (Vertically opposite angles), or
$
\angle B O D+\angle B O E=70^{\circ} .
$

But, we are given that $\angle B O D=40^{\circ}$.
$
\begin{aligned}
& 40^{\circ}+\angle B O E=70^{\circ} \\
& \angle B O E=70^{\circ}-40^{\circ} \\
& =30^{\circ} .
\end{aligned}
$

Therefore, we can conclude that Reflex $\angle C O E=250^{\circ}$ and $\angle B O E=30^{\circ}$.

Ex 6.1 Question 2.

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

Answer.

We are given that $\angle P O Y=90^{\circ}$ and $a: b=2: 3$.

We need find the value of $c$ in the given figure.

Let $a$ be equal to $2 x$ and $b$ be equal to $3 x$.
$
\begin{aligned}
& \because a+b=90^{\circ} \Rightarrow 2 x+3 x=90^{\circ} \Rightarrow 5 x=90^{\circ} \\
& \Rightarrow x=18^{\circ}
\end{aligned}
$

Therefore $b=3 \times 18^{\circ}=54^{\circ}$

Now $b+c=180^{\circ}$ [Linear pair]
$
\begin{aligned}
& \Rightarrow 54^{\circ}+c=180^{\circ} \\
& \Rightarrow c=180^{\circ}-54^{\circ}=126^{\circ}
\end{aligned}
$

Ex 6.1 Question 3.

$\text {In the given figure, } \angle P Q R=\angle P R Q \text {, then prove that } \angle P Q S=\angle P R T \text {. }$

Answer.

We need to prove that $\angle P Q S=\angle P R T$.

We are given that $\angle P Q R=\angle P R Q$.

From the given figure, we can conclude that $\angle P Q S$ and $\angle P Q R$, and $\angle P R S$ and $\angle P R T$ form a linear pair.
We know that sum of the angles of a linear pair is $180^{\circ}$.
$
\begin{aligned}
& \therefore \angle P Q S+\angle P Q R=180^{\circ} \text {, and (i) } \\
& \angle P R Q+\angle P R T=180^{\circ} \text {. (ii) }
\end{aligned}
$

From equations (i) and (ii), we can conclude that
$
\angle P Q S+\angle P Q R=\angle P R Q+\angle P R T
$

But, $\angle P Q R=\angle P R Q$.
$
\therefore \angle P Q S=\angle P R T \text {. }
$

Therefore, the desired result is proved.

Ex 6.1 Question 4.

$\text {In Fig. 6.16, if } x+y=w+z \text {, then prove that AOB is a line. }$

Answer.

We need to prove that $A O B$ is a line.

We are given that $x+y=w+z$.

We know that the sum of all the angles around a fixed point is $360^{\circ}$.

Thus, we can conclude that $\angle A O C+\angle B O C+\angle A O D+\angle B O D=360^{\circ}$, or
$
y+x+z+w=360^{\circ} .
$

But, $x+y=w+z$ (Given).
$
\begin{aligned}
& 2(y+x)=360^{\circ} . \\
& y+x=180^{\circ} .
\end{aligned}
$

From the given figure, we can conclude that $y$ and $x$ form a linear pair.

We know that if a ray stands on a straight line, then the sum of the angles of linear pair formed by the ray with respect to the line is $180^{\circ}$.
$
y+x=180^{\circ}
$

Therefore, we can conclude that $A O B$ is a line.

Ex 6.1 Question 5.

In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $\angle R O S=\frac{1}{2}(\angle Q O S-\angle P O S)$.

Answer.

We need to prove that
$
\angle R O S=\frac{1}{2}(\angle Q O S-\angle P O S)
$

We are given that $O R$ is perpendicular to $P Q$, or
$
\angle Q O R=90^{\circ} \text {. }
$

From the given figure, we can conclude that $\angle P O R$ and $\angle Q O R$ form a linear pair.

We know that sum of the angles of a linear pair is $180^{\circ}$.
$
\begin{aligned}
& \therefore \angle P O R+\angle Q O R=180^{\circ} \text {, or } \\
& \angle P O R=90^{\circ} .
\end{aligned}
$

From the figure, we can conclude that $\angle P O R=\angle P O S+\angle R O S$.
$
\begin{aligned}
& \Rightarrow \angle P O S+\angle R O S=90^{\circ} \text {, or } \\
& \angle R O S=90^{\circ}-\angle P O S \text {. }(l)
\end{aligned}
$

From the given figure, we can conclude that $\angle Q O S$ and $\angle P O S$ form a linear pair.

We know that sum of the angles of a linear pair is $180^{\circ}$.
$
\begin{aligned}
& \angle Q O S+\angle P O S=180^{\circ} \text {, or } \\
& \frac{1}{2}(\angle Q O S+\angle P O S)=90^{\circ}
\end{aligned}
$

Substitute (ii) in (i), to get
$
\begin{aligned}
& \angle R O S=\frac{1}{2}(\angle Q O S+\angle P O S)-\angle P O S \\
& =\frac{1}{2}(\angle Q O S-\angle P O S) .
\end{aligned}
$

Therefore, the desired result is proved.

Ex 6.1 Question 6.

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

Answer

 We are given that $\angle X Y Z=64^{\circ}, X Y$ is produced to $P$ and $Y Q$ bisects $\angle Z Y P$.

We can conclude the given below figure for the given situation:

We need to find $\angle X Y Q$ and reflex $\angle Q Y P$.

From the given figure, we can conclude that $\angle X Y Z$ and $\angle Z Y P$ form a linear pair.

We know that sum of the angles of a linear pair is $180^{\circ}$.
$
\begin{aligned}
& \angle X Y Z+\angle Z Y P=180^{\circ} \\
& \text { But } \angle X Y Z=64^{\circ} \\
& \Rightarrow 64^{\circ}+\angle Z Y P=180^{\circ} \\
& \Rightarrow \angle Z Y P=116^{\circ} .
\end{aligned}
$

Ray $Y Q$ bisects $\angle Z Y P$, or
$
\begin{aligned}
& \angle Q Y Z=\angle Q Y P=\frac{116^{\circ}}{2}= \\
& \angle X Y Q=\angle Q Y Z+\angle X Y Z \\
& =58^{\circ}+64^{\circ}=122^{\circ} .
\end{aligned}
$

Reflex $\angle Q Y P=360^{\circ}-\angle Q Y P$
$
\begin{aligned}
& =360^{\circ}-58^{\circ} \\
& =302^{\circ} .
\end{aligned}
$

$\text { Therefore, we can conclude that } \angle X Y Q=122^{\circ} \text { and Reflex } \angle Q Y P=302^{\circ}$