In the figure, $O$ is the centre of the circle. Find $\angle CBD$.
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Given:

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

To do:

We have to find $\angle CBD$.

Solution:

Arc $AC$ subtends $\angle AOC$ at the centre and $\angle APC$ at the remaining part of the circle.

This implies,

$\angle APC =\frac{1}{2}\angle AOC$

$= \frac{1}{2} \times 100^o$

$= 50^o$

$APCB$ is a.cyclic quadrilateral.

This implies,

$\angle APC + \angle ABC = 180^o$

$50^o + \angle ABC = 180^o$

$\angle ABC =180^o- 50^o$

$\angle ABC =130^o$

$\angle ABC + \angle CBD = 180^o$              (Linear pair)

$130^o + \angle CBD = 180^o$

$\angle CBD = 180^o- 130^o$

$= 50^o$

Hence $\angle CBD = 50^o$.

Tutorialspoint

Updated on: 10-Oct-2022

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