In figure, $O$ is the centre of the circle. If $\angle BAC=130^{\circ}$, then find $\angle BOC$."

Given: In the given figure, $O$ is the centre of the circle and $\angle BAC=130^{\circ}$.

To do: To find $\angle BOC$.

Solution:

Let us join $BC$. Now, let us take a point $P$ on the circumference of the circle and join $PB$ and $PC$.

$PBAC$ is a cyclic quadrilateral.

As known sum of opposite angles in a quadrilateral is $180^{\circ}$.

$\therefore \angle BAC+\angle BPC=180^{\circ}$

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

We know, $BC$ is a chord of the circle.

Angle subtended by chord at the centre of a circle is double of the angle subtended at it's circumference.

Therefore, $\angle BOC=2\angle BPC$

$\Rightarrow \angle BOC=50^{\circ}\times2=100^{\circ}$

Thus, $\angle BOC=100^{\circ}$.

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Updated on: 10-Oct-2022

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