Question

In the following figure QT$\perp$PR and QS$=$PS. If $\angle$TQR$={40}^o$ and $\angle$RPS$={20}^o$ then value of x is?

• A. ${80}^o$
• B. ${25}^o$
• C. ${15}^o$
• D. ${35}^o$

Answer 1

The correct option is C ${15}^o$

In $\Delta$PQS,
QS$=$PS
$\Rightarrow \angle$Q$=\angle$P$={40}^o+$x
In $\Delta$RQT,
$\angle$R$={180}^o-({90}^o+{40}^o)$
$\Rightarrow \angle$R$={50}^o$
In $\Delta$PQR,
${40}^o+x+{40}^o+x+{20}^o+{50}^o={180}^o$
$\Rightarrow x=\frac{{30}^o}{2}={15}^o$
Option $\left(3\right)$.