In Delta ABC,angle A = 2x - 10^{circ},angle B | vedclass

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Question

(A) The sum of the angles in a triangle is $180^{\circ}$.Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.Substituting the given values: $(2x

Vedclass · Accepted Answer

$\angle A = 70^{\circ}, \angle B = 50^{\circ}, \angle C = 60^{\circ}$

Vedclass · Answer

(A) The sum of the angles in a triangle is $180^{\circ}$.Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.Substituting the given values: $(2x - 10^{\circ}) + (x + 10^{\circ}) + (2x - 20^{\circ}) = 180^{\circ}$.Combining like terms: $5x - 20^{\circ} = 180^{\circ}$.$5x = 200^{\circ}$.$x = 40^{\circ}$.Now,calculate each angle:$\angle A = 2(40^{\circ}) - 10^{\circ} = 80^{\circ} - 10^{\circ} = 70^{\circ}$.$\angle B = 40^{\circ} + 10^{\circ} = 50^{\circ}$.$\angle C = 2(40^{\circ}) - 20^{\circ} = 80^{\circ} - 20^{\circ} = 60^{\circ}$.Thus,the angles are $\angle A = 70^{\circ}, \angle B = 50^{\circ}, \angle C = 60^{\circ}$.

In $\Delta ABC$,$\angle A = 2x - 10^{\circ}$,$\angle B = x + 10^{\circ}$,and $\angle C = 2x - 20^{\circ}$. Find the measure of each angle of $\Delta ABC$.

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