NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

Ans : AB is a straight line, rays OC and OE stand on it.
\(\begin{array}{l}{\therefore \angle \mathrm{AOC}+\angle \mathrm{COE}+\angle \mathrm{BOE}=180^{\circ}} \\ {\Rightarrow(\angle \mathrm{AOC}+\angle \mathrm{BOE})+\angle \mathrm{COE}=180^{\circ}} \\ {\Rightarrow 70^{\circ}+\angle \mathrm{COE}=180^{\circ}} \\ {\Rightarrow \angle \mathrm{COE}=180^{\circ}-70^{\circ}=110^{\circ}} \\ {\text { Reflex } \angle \mathrm{COE}=360^{\circ}-110^{\circ}=250^{\circ}}\end{array}\)
CD is a straight line. rays OE and 0B stand on it.
\(\begin{array}{l}{\therefore \angle \mathrm{COE}+\angle \mathrm{BOE}+\angle \mathrm{BOD}=180^{\circ}} \\ {\Rightarrow 110^{\circ}+\angle \mathrm{BOE}+40^{\circ}=180^{\circ}} \\ {\Rightarrow \angle \mathrm{BOE}=180^{\circ}-150^{\circ}=30^{\circ}}\end{array}\) 

Ans : Let the common ratio between a and b be x.
a = 2x, and b = 3x
XY is a straight line, rays OM and OP stand on it.
\(\begin{array}{l}{\therefore \angle \mathrm{XOM}+\angle \mathrm{MOP}+\angle \mathrm{POY}=180^{\circ}} \\ {b+a+\angle \mathrm{POY}=180^{\circ}} \\ {3 x+2 x+90^{\circ}=180^{\circ}} \\ {5 x=90^{\circ}} \\ {x=18^{\circ}} \\ {a=2 x=2 \times 18=36^{\circ}} \\ {b=3 x=3 \times 18=54^{\circ}}\end{array}\)
MN is a straight line. Ray OX stands on it.
\(\begin{array}{l}{\therefore b+c=180^{\circ}(\text { Linear Pair })} \\ {54^{\circ}+c=180^{\circ}} \\ {c=180^{\circ}-54^{\circ}=126^{\circ}} \\ {\therefore c=126^{\circ}}\end{array}\) 

Ans : In the given figure, ST isa straight line and ray QP stands on it.
\(\begin{array}{l}{\therefore \angle \mathrm{PQS}+\angle \mathrm{PQR}=180^{\circ}(\text { Linear Pair })} \\ {\angle \mathrm{PQR}=180^{\circ}-\angle \mathrm{PQS}(1)} \\ {\angle \mathrm{PRT}+\angle \mathrm{PRQ}=180^{\circ}(\text { Linear Pair })} \\ {\angle \mathrm{PRQ}=180^{\circ}-\angle \mathrm{PRT}(2)} \\ {\text { It is given that } \angle \mathrm{PQR}=\angle \mathrm{PRQ} \text { . }}\end{array}\)
Equating equations (1) and (2), we obtain
\(\begin{array}{l}{180^{\circ}-\angle \mathrm{PQS}=180^{\circ}-\angle \mathrm{PRT}} \\ {\angle \mathrm{PQS}=\angle \mathrm{PRT}}\end{array}\) 

Ans : \(\begin{array}{l}{\text { It can be observed that, }} \\ {x+y+z+w=360^{\circ}(\text { Complete angle })} \\ {\text { It is given that, }} \\ {x+y=z+w} \\ {\therefore x+y+x+y=360^{\circ}} \\ {2(x+y)=360^{\circ}} \\ {x+y=180^{\circ}}\end{array}\)
Since x and y form a linear pair, AOB is a line. 

Ans : It is given that \(\mathrm{OR} \perp \mathrm{PQ}\)
\(\begin{array}{l}{\square \square \mathrm{POR}=90^{\circ}} \\ {\square \square \mathrm{POS}+\square \mathrm{SOR}=90^{\circ}}\end{array}\)
\(\begin{array}{l}{\square R O S=90^{\circ}-\square P O S \ldots(1)} \\ {\square Q O R=90^{\circ}(A s O R \square P Q)}\end{array}\)
On adding equations (1) and (2), we obtain
\(2 \square \mathrm{ROS}=\square \mathrm{QOS}-\square \mathrm{POS}\)
\(\square \mathrm{ROS}=\frac{1}{2}(\square \mathrm{QOS}-\square \mathrm{POS})\)