Question

In $\Delta$PQR, seg $PQ$ $\cong$ seg $QR$ and $\angle P={70}^o$, then find the measures of angle $Q$ is:

Answer 1

Given, seg $PQ$ $\cong$ seg $QR$.
Thus, $\angle R=\angle P={70}^{\circ }$ (Isosceles triangle property).

We know, sum of angles of the triangle $={180}^o$.
Then, $\angle P+\angle Q+\angle R=180$
$⟹$ $70+70+\angle Q=180$
$⟹$ $\angle Q={40}^{\circ }$.