In figure below, $\angle PQR = 100^o$, where $P, Q$ and $R$ are points on a circle with centre $O$. Find $\angle OPR$.
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Given:

$\angle PQR = 100^o$, where $P, Q$ and $R$ are points on a circle with centre $O$.

To do:

We have to find $\angle OPR$.

Solution:

We know that,

The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the circle.

This implies,

Reflex $\angle POR = 2\angle PQR$

Reflex $\angle POR = 2\times100^o$

$= 200^o$

Therefore,

$\angle POR = 360^o-200^o$

$= 160^o$

In $\triangle OPR$,

$OP$ and $OR$ are the radii of the circle.

$OP = OR$

This implies,

$\angle OPR = \angle ORP$

$\angle POR+\angle OPR+\angle ORP = 180^o$

$\angle OPR+\angle OPR = 180^o-160^o$

$2\angle OPR = 20^o$

$\angle OPR=\frac{20^o}{2}$

$\angle OPR=10^o$

Hence, $\angle OPR = 10^o$.

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Updated on: 10-Oct-2022

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