Question

In the figure $△PQR$ is an isosceles triangle with PQ = PR and $\angle PQR={35}^{\circ }$. Find $\angle QTR$.

• A. ${40}^{\circ }$
• B. ${50}^{\circ }$
• C. ${60}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

The correct option is D ${70}^{\circ }$

In $△PQR$ :
$\angle PQR=\angle PRQ={35}^{\circ }$
[Angle opposite to equal sides]

Again, by angle sum property,
$\angle P+\angle Q+\angle R={180}^{\circ }$

$\angle P+{35}^{\circ }+{35}^{\circ }={180}^{\circ }$

$\angle P+{70}^{\circ }={180}^{\circ }$

$\angle P={180}^{\circ }-{70}^{\circ }$

$\angle P={110}^{\circ }$

Now, in quadrilateral SQTR,
$\angle QSR+\angle QTR={180}^{\circ }$
[Opposite angles of a cyclic quadrilateral]

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