Question

In fig. given below, $\angle ABC={69}^{\circ },\angle ACB={31}^{\circ }.$ Find $\angle BDC.$

Answer 1

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In $△ABC$, we have

$\angle A+\angle ABC+\angle ACB={180}^{\circ }$ [Angle sum property]

$\Rightarrow \angle A+{69}^{\circ }+31={180}^{\circ }$

$\Rightarrow \angle A={80}^{\circ }$

Since $\angle A$ & $\angle D$ are in same segment of circle.

$\angle A=\angle D={80}^{\circ }$ [Angles in the same segment of a circle are equal ]

$\therefore \angle BDC={80}^{\circ }$