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In figure below, $\angle ABC = 69^o, \angle ACB = 31^o$, find $\angle BDC$.
$\angle ABC = 69^o, \angle ACB = 31^o$
We have to find $\angle BDC$.
We know that,
Angles in the same segment of a circle are equal.
$\angle BAC = \angle BDC$
In $\triangle ABC$,
$\angle ABC+\angle BAC+\angle ACB = 180^o$ (The sum of the angles of a triangle is $180^o$)
$\angle BAC = 180^o-100^o$
$\angle BAC = 80^o$
$\angle BAC=\angle BDC = 80^o$
Hence, $\angle BDC = 80^o$.
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