Question

In the given figure, PQ = PR and $\angle PRQ={70}^{\circ }$. Find $\angle QAR$.

• A. ${40}^{\circ }$
• B. ${140}^{\circ }$
• C. ${70}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is A

${40}^{\circ }$

In $\Delta PQR$

PQ = PR (Given)

$\angle PQR=\angle PRQ={70}^{\circ }$ [Base angles of an isosceles triangle are equal]

So, $\angle QPR={180}^{\circ }-({70}^{\circ }+{70}^{\circ })={40}^{\circ }$

Also, $\angle QAR=\angle QPR$ [Angles in the same segment are equal].
$\therefore \angle QAR={40}^{\circ }$