In Delta ABC,AB = AC,angle A = 2x^{circ} and | vedclass

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(A) Given that in $\Delta ABC$,$AB = AC$. Therefore,the angles opposite to these sides are equal,i.e.,$\angle B = \angle C$.Since $\angle B = x + 10^{

Vedclass · Accepted Answer

$\angle A = 80^{\circ}, \angle B = 50^{\circ}, \angle C = 50^{\circ}$

Vedclass · Answer

(A) Given that in $\Delta ABC$,$AB = AC$. Therefore,the angles opposite to these sides are equal,i.e.,$\angle B = \angle C$.Since $\angle B = x + 10^{\circ}$,then $\angle C = x + 10^{\circ}$.The sum of all angles in a triangle is $180^{\circ}$.So,$\angle A + \angle B + \angle C = 180^{\circ}$.Substituting the given values: $2x + (x + 10) + (x + 10) = 180$.$4x + 20 = 180$.$4x = 160$.$x = 40$.Now,calculating the angles:$\angle A = 2x = 2(40) = 80^{\circ}$.$\angle B = x + 10 = 40 + 10 = 50^{\circ}$.$\angle C = x + 10 = 40 + 10 = 50^{\circ}$.

In $\Delta ABC$,$AB = AC$,$\angle A = 2x^{\circ}$ and $\angle B = x + 10^{\circ}$. Find each of the angles of $\Delta ABC$.

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