Suppose $O$ is the center of a circle and $ A B $ is a diameter of that circle. $ A B C D $ is a cyclic quadrilateral. If $ \angle A B C=65^{\circ}, \angle D A C=40^{\circ} $, then $ \angle B C D=? $

Given:

$O$ is the center of a circle and \( A B \) is a diameter of that circle. \( A B C D \) is a cyclic quadrilateral.

\( \angle A B C=65^{\circ}, \angle D A C=40^{\circ} \).

To do:

We have tofind \( \angle B C D \).

Solution:

We know that,

Angle in a semi-circle is $90^o$.

The sum of the opposite angles of a cyclic quadrilateral is $180^o$.

$\angle ACB$ is an angle in the semicircle.

This implies,

$\angle ACB = 90^o$

$\angle ABC$ and $\angle ADC$ are supplementary angles.

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

$65^o+\angle ADC = 180^o$

$\angle ADC = 180^o - 65^o$

$\angle ADC = 115^o$

In $\triangle ADC$,

$\angle ADC = 115^o, \angle DAC = 40^o$

$\angle ACD+\angle ADC+\angle DAC=180^o$

$\angle ACD = 180^o - \angle ADC - \angle DAC$

$= 180^o- 115^o- 40^o$

$= 25^o$

$\angle BCD = \angle ACB+\angle ACD$

$= 90^o+25^o$

$= 115^o$ 

Hence, $\angle BCD = 115^o$.

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Updated on: 10-Oct-2022

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