In the figure, $O$ is the centre of the circle. If $\angle BOD = 160^o$, find the values of $x$ and $y$.
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Given:

In the figure, $O$ is the centre of the circle.

$\angle BOD = 160^o$.

To do:

We have to find the values of $x$ and $y$.

Solution:

$ABCD$ is a cyclic quadrilateral.

Arc $BAD$ subtends $\angle BOD$ is the angle at the centre and $\angle BCD$ is on the other part of the circle.

Therefore,

$\angle BCD = \frac{1}{2}\angle BOD$

$x = \frac{1}{2} \times 160^o = 80^o$

$ABCD$ is a cyclic quadrilateral.

This implies,

$\angle A + \angle C = 180^o$

$y + x = 180^o$

$y + 80^o = 180^o$

$y =180^o- 80^o = 100^o$

Hence, $x = 80^o$ and $y = 100^o$.

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Updated on: 10-Oct-2022

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