Question

Question 30

In the given figure, lines PQ and ST intersect at O. If $\angle POR={90}^{\circ }$and
x:y = 3:2, then z is equal to

(a) ${126}^{\circ }$
(b) ${144}^{\circ }$
(c) ${136}^{\circ }$
(d) ${154}^{\circ }$

Answer 1

Since, $\angle POR,\angle ROT$ and $\angle TOQ$
lies on a straight line POQ, then their sum is equal to ${180}^{\circ }$
$\therefore \angle POR+\angle ROT+\angle TOQ={180}^{\circ }\Rightarrow {90}^{\circ }+x+y={180}^{\circ }\to x+y={180}^{\circ }-{90}^{\circ }\Rightarrow x+y={90}^{\circ }Alsox:y=3:2$
Let $x=3aandy=2a\therefore 3a+2a={90}^{\circ }\Rightarrow 5a={90}^{\circ }\Rightarrow a=\frac{{90}^{\circ }}{5}={18}^{\circ }$
Now, $x=3a=3\times {18}^{\circ }={54}^{\circ }andy=2a=2\times {18}^{\circ }={36}^{\circ }$
Since, y and z forms a linear pair,
$\therefore y+z={180}^{\circ }\Rightarrow {36}^{\circ }+z={180}^{\circ }\Rightarrow {180}^{\circ }-{36}^{\circ }\Rightarrow z={144}^{\circ }$