Question

The given figure shows a circle with centre $O$. Also, $PQ=QR=RS$ and $\angle PTS={75}^o$. Calculate:
i) $\angle POS$,
ii) $\angle QOR$,
iii) $\angle PQR$.

Answer 1

$PQ=QR=RS$

$\Rightarrow \angle PTS=75°$

$i).\angle POS=2\angle PTS$

$=2\times 75°$

$=150°$

$ii).\angle QOR=\frac{1}{3}\times \angle POS=\frac{1}{3}\times 150°=50°$

$\because PO=QO=RO=SO=$ radius

$\therefore \angle OPQ=\angle OQP=\angle ORQ=\angle ORS=\angle OSR$ and hence $\angle POQ=\angle QOR=\angle ROS$

$iii).\angle PQR$

In $△POQ$

$\Rightarrow \angle POQ=50°$

$\angle OPQ=\angle OQP$

$\Rightarrow 50°+2\angle OQP=180°$

$\Rightarrow \angle OQP=\frac{130°}{2}=65°$

similarly $\angle OQR=65°$

$\therefore \angle PQR=2\times 65°=130°.$

Hence, solved.