Question

In the given figure, line $PQ$ 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 $\angle BOQ$.

• A. $\angle SOQ={40}^o$
• B. $\angle SOQ={110}^o$
• C. $\angle BOQ={110}^o$
• D. $\angle BOQ={40}^o$

Answer 1

Answer: (A), (C)

$\angle BOQ={110}^o$
$\angle SOQ={40}^o$

$\angle POR+\angle POS={180}^o$ .....(1) (Linear pair of angles)
Ir is given that, ray OA and ray OB bisect $\angle POR$ and $\angle POS$ respectively.
Therefore,
$\angle POA=\frac{1}{2}\angle POR$ and $\angle POB=\frac{1}{2}\angle POS$
$\Rightarrow \angle POA+\angle POB=\frac{1}{2}\{POR+POS\}$
$=\frac{1}{2}\times {180}^o={90}^o$ (From equation 1)
Now, if $\angle POA+\angle POB=2:7$
Sum of the ratios $=2+7=9$
then, we have $\angle POA=\frac{2}{9}\times {90}^o={20}^o$
and $\angle POB=\frac{7}{9}\times {90}^o={70}^o$
$\angle POR=2\times \angle POA=2\times {20}^o={40}^o$
$\angle SOQ=\angle POR$ (Vertically opposite angles)
$\therefore \angle SOQ={40}^o$
$\angle BOQ=\angle BOS+\angle SOQ$
$=\angle POB+\angle SOQ$ (Since, OB bisect $\angle POS)$
$(\angle BOS=\angle POB)$
$={70}^o+{40}^o={110}^o$
$\therefore \angle BOQ={110}^o$