In Delta ABC,angle B = frac{angle A + angle C | vedclass

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Question

(A) We know that the sum of angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.Given $\angle B = \frac{\angle A +

Vedclass · Accepted Answer

$\angle A = 40^{\circ}, \angle B = 60^{\circ}, \angle C = 80^{\circ}$

Vedclass · Answer

(A) We know that the sum of angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.Given $\angle B = \frac{\angle A + \angle C}{2}$,which implies $2\angle B = \angle A + \angle C$.Substituting this into the sum equation: $2\angle B + \angle B = 180^{\circ} \implies 3\angle B = 180^{\circ} \implies \angle B = 60^{\circ}$.Now,$\angle A + \angle C = 180^{\circ} - 60^{\circ} = 120^{\circ}$.Given the ratio $\angle A : \angle C = 1 : 2$,let $\angle A = x$ and $\angle C = 2x$.Then $x + 2x = 120^{\circ} \implies 3x = 120^{\circ} \implies x = 40^{\circ}$.Therefore,$\angle A = 40^{\circ}$ and $\angle C = 2(40^{\circ}) = 80^{\circ}$.The angles are $\angle A = 40^{\circ}, \angle B = 60^{\circ}, \angle C = 80^{\circ}$.

In $\Delta ABC$,$\angle B = \frac{\angle A + \angle C}{2}$ and $\angle A : \angle C = 1 : 2$. Find the measure of each angle of $\Delta ABC$.

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