Question

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

• A. ${20}^o$
• B. ${60}^o$
• C. ${30}^o$
• D. ${80}^o$

Answer 1

Answer: (B)

${60}^o$

In $△PRT$

$\angle PTS=\angle PRT+\angle RTP$ (As exterior angle is equal to sum of interior opposite angles )

$\Rightarrow \angle PTS={95}^{\circ }+{40}^{\circ }={135}^{\circ }$

Now in $△SQT$

$\angle STQ=\angle SQT+\angle TSQ$ (As exterior angle is equal to sum of interior opposite angles )

$\Rightarrow {135}^{\circ }=\angle SQT+{75}^{\circ }\Rightarrow \angle SQT={135}^{\circ }-{75}^{\circ }={60}^{\circ }$