Question

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

Answer 1

In $\Delta PRT$

$\Rightarrow \angle PTR+\angle PRT+\angle RPT={180}^{\circ }$ (since the sum of the angles of a triangle is ${180}^{\circ }$)

$\Rightarrow \angle PTR+{40}^{\circ }+{95}^{\circ }={180}^{\circ }$

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

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

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

$\Rightarrow \angle QTS=\angle PTR={45}^{\circ }$ (Vertically Opposite angles)

In $\Delta TSQ$

$\Rightarrow \angle QTS+\angle TSQ+\angle SQT={180}^{\circ }$ (since the sum of the angles of a triangle is ${180}^{\circ }$)

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

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

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

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