Question

$\angle PQR={100}^{\circ }$, where $P,Q$ and $R$ are points on a circle with centre $O$. Find $\angle OPR$

Answer 1

Here, PR is chord

We mark s on major arc of the circle.

$\therefore$ PQRS is a cyclic quadrilateral.

So, $\angle PQR+\angle PSR={180}^o$

[Sum of opposite angles of a cyclic quadrilateral is ${180}^o$]

$100+\angle PSR={180}^o$

$\angle PSR={180}^o-{100}^o$

$\angle PSR={80}^o$

Arc PQR subtends $\angle PQR$ at centre of a circle.

And $\angle PSR$ on point s.

So, $\angle POR=2\angle PSR$

[Angle subtended by arc at the centre is double the angle subtended by it any other point]

$\angle POR=2\times {80}^o={160}^o$

Now,

In $\Delta$OPR,

$OP=OR$[Radii of same circle are equal]

$\therefore \angle OPR=\angle ORP$ [opp. angles to equal sides are equal] ………………..$\left(1\right)$

Also in $\Delta OPR$,

$\angle OPR+\angle ORP+\angle POR={180}^o$ (Angle sum property of triangle)

$\angle OPR+\angle OPR+\angle POR={180}^o$ from $\left(1\right)$

$2\angle OPR+160={180}^o$

$2\angle OPR={180}^o-{160}^o$

$2\angle OPR=20$

$\angle OPR=20/2$

$\therefore \angle OPR={10}^o$.