Question

In the figure, lines PQ and RS intersect each other at point O; ray OA and ray OB bisect $\angle POR$ and $\angle POS$ respectively. If $\angle POA:\angle POB=2:7$, then find $\angle SOQ$ and $\angle BOQ$. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 2 Marks
Answer: 0.5 Mark each

$\angle POR+\angle POS={180}^{\circ }$ ...(1) (Linear pair of angles)

It is given that ray OA and ray OB bisect $\angle POR$ and $\angle POS$ respectively.

Therefore, $\angle POA=\frac{1}{2}\angle POR$

$\angle POB=\frac{1}{2}\angle POS$

$\angle POA+\angle POB=\frac{1}{2}\{POR+POS\}$

$=\frac{1}{2}\times {180}^{\circ }={90}^{\circ }$

$\angle POA+\angle POB=2:7$

Sum of the ratios = 2+7 = 9

$\angle POA=\frac{2}{9}\times {90}^{\circ }={20}^{\circ }$

$\angle POB=\frac{7}{9}\times {90}^{\circ }={70}^{\circ }$

$\angle POR=2\times \angle POA=2\times {20}^{\circ }={40}^{\circ }$

$\angle SOQ=\angle POR$ (Vertically opposite angles)

$\therefore \angle SOQ={40}^{\circ }$

$\angle BOQ=\angle BOS+\angle SOQ$

$=\angle POB+\angle SOQ$

$={70}^{\circ }+{40}^{\circ }={110}^{\circ }$

$\therefore \angle BOQ={110}^{\circ }$