Question

In the figure, $\Delta PQR$ is an isosceles triangle with PQ = PR and m $\angle PQR={35}^{\circ }$, Find m $\angle QSR$ and m $\angle QTR$

Answer 1

In the figure, $\Delta PQR$ is an isosceles triangle with PQ = PR.

$\angle PQR={35}^{\circ }$

$\therefore PRQ={35}^{\circ }$

But $\angle PQR+\angle PRQ+\angle QPR={180}^{\circ }$ (Sum of angles of a triangle)

$\Rightarrow {35}^{\circ }+{35}^{\circ }+\angle QPR={180}^{\circ }$

$\Rightarrow {70}^{\circ }+\angle QPR={180}^{\circ }$

$\Rightarrow \angle QPR={180}^{\circ }-{70}^{\circ }={110}^{\circ }$

$\because \angle QSR=\angle QPR$ (Angles in the same segment of circles)

$\therefore \angle QSR={110}^{\circ }$

But PQTR is a cyclic quadrilateral.

$\therefore \angle QTR+\angle QPR={180}^{\circ }$

$\Rightarrow \angle QTR+{110}^{\circ }={180}^{\circ }$

$\Rightarrow \angle QTR={180}^{\circ }-{110}^{\circ }={70}^{\circ }$

Hence, $\angle QTR={70}^{\circ }$