Question

$QP\parallel TS$ and $\angle QRS={36}^o$. Calculate $\angle PQR$.

• A. ${75}^o$
• B. ${70}^o$
• C. ${72}^o$
• D. ${71}^o$

Answer 1

The correct option is C ${72}^o$

Construction:
Extend Ray TS to QR at O such that $m\angle$SOQ$=$$m\angle$PQO.
Solution:
$\therefore$ $m\angle$$SOR$$=$${180}^o-X^o$
Also, $m\angle$$RSO$$=$${180}^o-2X^o$
Since, $m\angle$$SOR$+ $m\angle$$RSO$+ $m\angle$$SRO$$=$${180}^o$

$\therefore$${180}^o{-}^o+{180}^o-2X^o+{36}^o$$=$${180}^o$

$\therefore X$=${72}^o$

$\therefore$$\angle PQR$$=$${72}^o$