Question

In fig, if lines PQ and RS intersect at a point T such that $\angle PRT={40}^{\circ }$,$\angle RPT={95}^{\circ }$ and $\angle TSQ={75}^{\circ }$, find $\angle SQT$.

Answer 1

In PRT

By angle sum property of triangle

$\angle PRT+\angle RPT+\angle PTR={180}^{\circ }$

$\Rightarrow {90}^{\circ }+{45}^{\circ }+\angle PTR={180}^{\circ }$

$\Rightarrow {135}^{\circ }+\angle PTR={180}^{\circ }$

$\Rightarrow \angle PTR={180}^{\circ }-{135}^{\circ }$

$\Rightarrow \angle PTR={45}^{\circ }$.

Also,

$\angle STQ=\angle PTR$

$\angle STQ={45}^{\circ }$ (vertically opposite angles)

In $\Delta SQT$

By angle sum property of triangle

$\angle SQT+\angle STQ+\angle TSQ={180}^{\circ }$

$\angle SQT+{75}^{\circ }+{45}^{\circ }={180}^{\circ }$

$\angle SQT+{120}^{\circ }={180}^{\circ }$

$\angle SQT={180}^{\circ }-{120}^{\circ }$

$\angle SQT={60}^{\circ }$