Question

Question 17

In $\Delta PQR$, if PQ = QR and $\angle Q={100}^{\circ }$, then $\angle R$ is equal to

a) ${40}^{\circ }$
b) ${80}^{\circ }$
c) ${120}^{\circ }$
d) ${50}^{\circ }$

Answer 1

In $\Delta PQR$, PQ = QR [given]
Let $\angle P=x$
Therefore,
$\angle$R = x [since equal sides opposite angles are equal]

$\angle P+\angle Q+\angle R={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow x+{100}^{\circ }+x={180}^{\circ }$
$\Rightarrow 2x+{100}^{\circ }={180}^{\circ }\Rightarrow 2x={80}^{\circ }\Rightarrow x={40}^{\circ }$
Hence, $\angle P=\angle R={40}^{\circ }$