In figure below, $\angle ABC = 69^o, \angle ACB = 31^o$, find $\angle BDC$.
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Given:

$\angle ABC = 69^o, \angle ACB = 31^o$

To do:

We have to find $\angle BDC$.

Solution:

We know that,

Angles in the same segment of a circle are equal.

This implies,

$\angle BAC = \angle BDC$

In $\triangle ABC$,

$\angle ABC+\angle BAC+\angle ACB = 180^o$       (The sum of the angles of a triangle is $180^o$)

$69^o+\angle BAC+31^o=180^o$

$\angle BAC = 180^o-100^o$

$\angle BAC = 80^o$

This implies,

$\angle BAC=\angle BDC = 80^o$

Hence, $\angle BDC = 80^o$.

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Updated on: 10-Oct-2022

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