Question

In the figure, OA and OB are opposite rays:
(i) If x=${25}^{\circ }$, what is the value of y?
(ii) If y=${35}^{\circ }$, what is the value of x?

Answer 1

(i) If x = ${25}^{\circ }$
$\therefore 3x=3\times {25}^{\circ }={75}^{\circ }$
But $\angle AOC+\angle BOC={180}^{\circ }$ (Linear pair)
$\Rightarrow \angle AOC+{75}^{\circ }={180}^{\circ }\Rightarrow \angle AOC={180}^{\circ }-{75}^{\circ }\Rightarrow \angle AOC={105}^{\circ }\therefore 2y+5={105}^{\circ }\Rightarrow 2y={105}^{\circ }-5^{\circ }={100}^{\circ }\Rightarrow y=\frac{{100}^{\circ }}{2}={50}^{\circ }$
$\therefore$ If x=${25}^{\circ }$ they y=${50}^{\circ }$
(ii) If y= ${35}^{\circ }$, then $\angle AOC=2y+5\therefore 2y+5=2\times {35}^{\circ }+5={70}^{\circ }+5={75}^{\circ }$
But $\angle AOC+\angle BOC={180}^{\circ }$ (Linear pair)
$\Rightarrow {75}^{\circ }+\angle BOC={180}^{\circ }\Rightarrow \angle BOC={180}^{\circ }-{75}^{\circ }={105}^{\circ }\therefore 3x={105}^{\circ }\Rightarrow x=\frac{{105}^{\circ }}{3}={35}^{\circ }\therefore x={35}^{\circ }$