Question

In the following figure, if lines PQ and RS intersect at point T, such that $\angle$PRT = 40°, $\angle$RPT = $\angle$95° and $\angle$TSQ = 75°, find $\angle$SQT.

Answer 1

In $∆PRT$
$$\begin{gathered} \angle RPT+\angle PRT+\angle RTP={180}^{\circ }\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow {95}^{\circ }+{40}^{\circ }+\angle RTP={180}^{\circ } \\ \Rightarrow \angle RTP={180}^{\circ }-{135}^{\circ }={45}^{\circ } \end{gathered}$$
$\angle RTP\mathrm{and}\angle STQ$ are vertically opposite angles, so
$\angle STQ=\angle RTP={45}^{\circ }$
Now, in $∆STQ$
$$\begin{gathered} \angle STQ+\angle TSQ+\angle SQT={180}^{\circ }\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow {45}^{\circ }+{75}^{\circ }+\angle SQT={180}^{\circ } \\ \Rightarrow \angle SQT={180}^{\circ }-{120}^{\circ }={60}^{\circ } \end{gathered}$$
Hence, $\angle$SQT = 60°.