Question

In the figure, if lines $PQ$ 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

Given, $\angle PRT={40}^{\circ },\angle RPT={95}^{\circ },\angle TSQ={75}^{\circ }$
According to the question,
$\angle PRT+\angle RPT+\angle PTR={180}^{\circ }$ (Sum of the interior angles of the triangle.)
$\Rightarrow {40}^{\circ }+{95}^{\circ }+\angle PTR={180}^{\circ }$
$\Rightarrow {135}^{\circ }+\angle PTR={180}^{\circ }$
$\Rightarrow \angle PTR={45}^{\circ }$
$\angle PTR=\angle STQ={45}^{\circ }$ (Vertically opposite angles.)
Now, $\angle TSQ+\angle PTR+\angle SQT={180}^{\circ }$(Sum of the interior angles of the triangle.)

${75}^{\circ }+{45}^{\circ }+\angle SQT={180}^{\circ }$
$\Rightarrow {120}^{\circ }+\angle SQT={180}^{\circ }$
$\Rightarrow \angle SQT={60}^{\circ }$