Question

If $\angle PQR={65}^o,\angle QPT={60}^o,\angle SPR={35}^o$, find the value of:
(i) $\angle QPR$ (ii) $\angle PRS$ (iii) $\angle PSR$ (iv) $\angle PQT$.

Answer 1

$\angle QPR=?$

$\angle PRS=?$

$\angle PSR=?$

$\angle PQT=?$

In $\Delta PRQ$,

$\angle P+\angle R+\angle Q={180}^{\circ }$ ...(Angle sum property)

$⟹$ $\angle QPR+{90}^{\circ }+{65}^{\circ }={180}^{\circ }$

$⟹$ $\angle QPR={180}^{\circ }-{155}^{\circ }={25}^{\circ }$.

In $\Delta PQT$,

$\angle P+\angle Q+\angle T={180}^{\circ }$ ...(Angle sum property)

$⟹$ $\angle PQT+{90}^o+{60}^o={180}^o$

$⟹$ $\angle PQT={180}^{\circ }-{150}^{\circ }={30}^{\circ }$.

Now,

$\angle PSR+\angle PQR={180}^{\circ }$ ...(Sum of opposite sides of cyclic quad= ${180}^{\circ }$)

$⟹$ $\angle PSR={180}^{\circ }-{65}^{\circ }={115}^{\circ }$.

In $\Delta PSR$,

$\angle PSR+\angle P+\angle PRS={180}^{\circ }$ ...(Angle sum property)

$⟹$ $\angle PSR={180}^{\circ }-{35}^{\circ }-{115}^{\circ }={180}^{\circ }-{150}^{\circ }$

$⟹$ $\angle PRS={30}^{\circ }$.