Lines and Angles

Class 9 NCERT Maths

NCERT

1   In the given figure, lines $AB$ and $CD$ intersect at $O$. If $\angle AOC +\angle BOE = 70^o$ and $\angle BOD =40^o$ find $\angle BOE$ and reflex $\angle COE.$

$AB$ is a straight line, rays $OC$ and $OE$ stand on it.$\\$ $\therefore \angle AOC+\angle COE+\angle BOE=180^o\\ \Rightarrow (\angle AOC+\angle BOE)+\angle COE=180^o\\ \Rightarrow 70^o+\angle COE=180^o\\ \Rightarrow \angle COE=180^o-70^o=110^o$$\\ Reflex \angle COE=360^o-110^o=250^o$$\\$ $CD$ is a straight line, rays $OE$ and $OB$ stand on it.$\\$ $\therefore \angle COE+\angle BOE+\angle BOD=180^o\\ \Rightarrow 110^o+\angle BOE+40^o=180^o\\ \Rightarrow \angle BOE=180^o-150^o=30^o$$\\ Hence, \angle BOE = 30^o and Reflex \angle COE = 250^o 2 In the given figure, lines XY and MN intersect at O. If \angle POY = 90^o and a: b = 2:3, find c. Solution : Let the common ratio between a and b be x.$$\\$ $\therefore a=2x$ and $b=3x$$\\ XY is a straight line, rays OM and OP stand on it.\\ \therefore \angle XOM+\angle MOP+\angle POY=180^o$$\\$ $b+a+\angle POY =180^o$$\\ 3x+2x+90^o=180^o$$\\$ $5x=90^o$$\\ x=18^o$$\\$ $a=2x=2*18=36^o$$\\ b=3x=3*18=54^o$$\\$ $MN$ is a straight line. Ray $OX$ stands on it.$\\$ $\therefore b+c=180^o$(Linear Pair)$\\$ $54^o+c=180^o$$\\ c=180^o-54^o=126^o$$\\$ $\therefore c=126^o$

3   In the given figure, $\angle PQR = \angle PRQ,$ then prove that $\angle PQS = \angle PRT.$

Solution :

In the given figure, $ST$ is a straight line and ray $QP$ stands on it.$\\$ $\therefore \angle PQS + \angle PQR = 180^o$(Linear Pair)$\\$ $\angle PQR = 180^o - \angle PQS ... (1)$$\\ \angle PRT + \angle PRQ = 180^o (Linear Pair)\\ \angle PRQ = 180^o - \angle PRT ... (2)$$\\$ It is given that $\angle PQR = \angle PRQ.$$\\ Equating Equations (1) and (2), we obtain\\ 180^o -\angle PQS = 180 - \angle PRT$$\\$ $\angle PQS = \angle PRT$(Proved)

4   In the given figure, if $x + y = w + z$ then prove that $AOB$ is a line.

Solution :

It can be observed that,$\\$ $x + y + z + w = 360^o$ (Complete angle)$\\$ It is given that,$\\$ $x + y = z + w$$\\ \therefore x + y + x + y = 360^o$$\\$ $2(x + y) = 360^o$$\\ x + y = 180^o$$\\$ Since $x$ and $y$ form a linear pair, therefore, $AOB$ is a line. (Proved)

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

Solution :

It is given that $OR \perp PQ$$\\ \angle POR = 90^o\\ \angle POS + \angle SOR = 90^o\\ \angle ROS = 90^o - \angle POS ... (1)\\ \angle QOR = 90^o (As \ OR \perp PQ)\\ \angle QOS - \angle ROS = 90^o\\ \angle ROS = \angle QOS - 90^o ... (2)$$\\$ On adding Equations (1) and (2), we obtain$\\$ $2 \angle ROS = \angle QOS - \angle POS$$\\$ $\angle ROS=\dfrac{1}{2}(\angle QOS-\angle POS)$