Question

From the adjacent figure find the value of (i) $\angle PRS$ (ii) $\angle PTS$ (iii) $\angle STR$ (iv) $\angle PRQ$

Answer 1

From $△PQR,\angle RPQ+\angle PQR+\angle PRQ={180}^{\circ }$ using angle sum property,

$\Rightarrow {50}^{\circ }+{35}^{\circ }+\angle PRQ={180}^{\circ }$

$\Rightarrow \angle PRQ={180}^{\circ }-{85}^{\circ }={95}^{\circ }$

As $QS$ is a straight line, $\angle QRS={180}^{\circ }$

$\Rightarrow \angle QRS=\angle QRT+\angle TRS$

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

$\Rightarrow \angle TRS={180}^{\circ }-{95}^{\circ }={85}^{\circ }$

In $△TRS,\angle TRS+\angle RST+\angle STR={180}^{\circ }$

$\Rightarrow {85}^{\circ }+{45}^{\circ }+\angle STR={180}^{\circ }$

$\Rightarrow \angle STR={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Again $PR$ is a straight line and $R$ is a point on the line $PR$

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

$\Rightarrow \angle PTS+\angle STR=\angle PTR$

$\Rightarrow \angle PTS+{50}^{\circ }={180}^{\circ }$

$\therefore \angle PTS={180}^{\circ }-{50}^{\circ }={130}^{\circ }$

Hence

$\left(i\right)\angle PRS=\angle TRS={85}^{\circ }$

$\left(ii\right)\angle PTS={130}^{\circ }$

$\left(iii\right)\angle STR={50}^{\circ }$

$\left(iv\right)\angle PRQ={95}^{\circ }$