In the figure, $O$ is the centre of the circle. If $\angle CEA = 30^o$, find the values of $x, y$ and $z$.
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Given:

$O$ is the centre of the circle.

$\angle CEA = 30^o$

To do:

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

Solution:

$\angle AEC$ and $\angle ADC$ are in the same segment.

Therefore,

$\angle AEC = \angle ADC = 30^o$

$z = 30^o$

$ABCD$ is a cyclic quadrilateral.

This implies,

$\angle B + \angle D = 180^o$

$x + z = 180^o$

$x + 30^o = 180^o$

$x = 180^o - 30^o = 150^o$

Arc $AC$ subtends $\angle AOB$ at the centre and $\angle ADC$ at the remaining part of the circle.

This implies,

$\angle AOC = 2\angle D$

$= 2 \times 30^o$

$= 60^o$

$y = 60^o$

Hence $x = 150^o, y = 60^o$ and $z = 30^o$.

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Updated on: 10-Oct-2022

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