In Delta ABC,angle B = 70^{circ} and angle C | vedclass

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(B) In $\Delta ABC$,the sum of angles is $180^{\circ}$.$\angle A + \angle B + \angle C = 180^{\circ}$$\angle A + 70^{\circ} + 80^{\circ} = 180^{\circ}

Vedclass · Accepted Answer

$95^{\circ}$ and $85^{\circ}$

Vedclass · Answer

(B) In $\Delta ABC$,the sum of angles is $180^{\circ}$.$\angle A + \angle B + \angle C = 180^{\circ}$$\angle A + 70^{\circ} + 80^{\circ} = 180^{\circ}$$\angle A = 180^{\circ} - 150^{\circ} = 30^{\circ}$.Since $AD$ is the bisector of $\angle A$,$\angle BAD = \angle CAD = \frac{30^{\circ}}{2} = 15^{\circ}$.In $\Delta ABD$,$\angle ADB = 180^{\circ} - (\angle B + \angle BAD) = 180^{\circ} - (70^{\circ} + 15^{\circ}) = 180^{\circ} - 85^{\circ} = 95^{\circ}$.In $\Delta ADC$,$\angle ADC = 180^{\circ} - (\angle C + \angle CAD) = 180^{\circ} - (80^{\circ} + 15^{\circ}) = 180^{\circ} - 95^{\circ} = 85^{\circ}$.

In $\Delta ABC$,$\angle B = 70^{\circ}$ and $\angle C = 80^{\circ}$. The bisector of $\angle A$ intersects $BC$ at $D$. Find $\angle ADB$ and $\angle ADC$.

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