In Delta ABC,angle A + angle B = 80^{circ} an | vedclass

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

Vedclass · Accepted Answer

$\angle A = 30^{\circ}, \angle B = 50^{\circ}, \angle C = 100^{\circ}$

Vedclass · Answer

(A) We know that the sum of all angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.Given: $\angle A + \angle B = 80^{\circ}$ and $\angle B + \angle C = 150^{\circ}$.Substituting $\angle A + \angle B = 80^{\circ}$ into the sum equation: $80^{\circ} + \angle C = 180^{\circ} \implies \angle C = 180^{\circ} - 80^{\circ} = 100^{\circ}$.Substituting $\angle C = 100^{\circ}$ into $\angle B + \angle C = 150^{\circ}$: $\angle B + 100^{\circ} = 150^{\circ} \implies \angle B = 50^{\circ}$.Substituting $\angle B = 50^{\circ}$ into $\angle A + \angle B = 80^{\circ}$: $\angle A + 50^{\circ} = 80^{\circ} \implies \angle A = 30^{\circ}$.Thus,the angles are $\angle A = 30^{\circ}, \angle B = 50^{\circ}, \angle C = 100^{\circ}$.

In $\Delta ABC$,$\angle A + \angle B = 80^{\circ}$ and $\angle B + \angle C = 150^{\circ}$. Find the measure of each angle of $\Delta ABC$.

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