Question

Find the value of $x$ in degree. $SQ$ is the angle bisector of $\angle PSR$ and $P$ & $R$ are perpendiculars on side $SP$ and $SR$ from $Q.$

• A. ${10}^o$
• B. ${11}^o$
• C. ${12}^o$
• D. ${13}^o$

Answer 1

The correct option is C ${12}^o$

Since, ray SQ is the bisector of $\angle PSR$,

$\therefore \angle PSQ=\angle RSQ$

$\therefore (9x-15{)}^o=(4x+45{)}^o$
$\therefore 9x-4x=45+15$

$\therefore 5x={60}^o$

$\therefore x={12}^o$

So, option $C$ is correct.