angle ACD is an exterior angle of Delta ABC. | vedclass

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(D) In $\Delta ABC$,we are given that $AB = AC$. According to the property of an isosceles triangle,angles opposite to equal sides are equal. Therefor

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(D) In $\Delta ABC$,we are given that $AB = AC$. According to the property of an isosceles triangle,angles opposite to equal sides are equal. Therefore,$\angle C = \angle B = 70^{\circ}$. The sum of angles in a triangle is $180^{\circ}$. So,$\angle A + \angle B + \angle C = 180^{\circ}$. $\angle A + 70^{\circ} + 70^{\circ} = 180^{\circ}$. $\angle A + 140^{\circ} = 180^{\circ}$,which gives $\angle A = 40^{\circ}$. The exterior angle property states that the exterior angle of a triangle is equal to the sum of the two interior opposite angles. Thus,$\angle ACD = \angle A + \angle B$. $\angle ACD = 40^{\circ} + 70^{\circ} = 110^{\circ}$.

$\angle ACD$ is an exterior angle of $\Delta ABC$. If $AB = AC$ and $\angle B = 70^{\circ}$,then $\angle ACD = \dots$ (in $^{\circ}$)

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