Question

In the given 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

Using angle sum property for $△PRT$, we obtain,
$\angle PRT+\angle RPT+\angle PTR={180}^{\circ }$
${40}^{\circ }+{95}^{\circ }+\angle PTR={180}^{\circ }$
$\angle PTR={180}^{\circ }-{135}^{\circ }$
$\angle PTR={45}^{\circ }$
$\angle STQ=\angle PTR={45}^{\circ }$ (vertically opposite angles)
$\angle STQ={45}^{\circ }$
By using angle sum property for $△STQ$, we obtain,
$\angle STQ+\angle SQT+\angle QST={180}^{\circ }$
${45}^{\circ }+\angle SQT+{75}^{\circ }={180}^{\circ }$
$\angle SQT={180}^{\circ }-{120}^{\circ }$
$\angle SQT={60}^{\circ }$