Question

In figure, if QT $\perp$ PR, $\angle$ TQR=${40}^o$ and $\angle$SPR$={30}^o$, then y is

• A. ${70}^o$
• B. ${110}^o$
• C. ${90}^o$
• D. ${80}^o$

Answer 1

The correct option is D ${80}^o$

$\angle MTP+\angle TPM=\angle TMS$
$30+90=\angle TMS$
$\angle TMS={120}^o$
$\angle Q+\angle M+\angle S={180}^o$
$y={180}^o-\left(100\right)$
$y={180}^o$