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

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(C) According to the Exterior Angle Theorem of a triangle,the measure of an exterior angle of a triangle is equal to the sum of the measures of its tw

Vedclass · Accepted Answer

$115$

Vedclass · Answer

(C) According to the Exterior Angle Theorem of a triangle,the measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.In $\Delta ABC$,$\angle ACD$ is the exterior angle at vertex $C$.The interior opposite angles are $\angle A$ and $\angle B$.Therefore,$\angle ACD = \angle A + \angle B$.Given $\angle A = 50^{\circ}$ and $\angle B = 65^{\circ}$.Substituting these values,we get $\angle ACD = 50^{\circ} + 65^{\circ} = 115^{\circ}$.

$\angle ACD$ is an exterior angle of $\Delta ABC$. If $\angle A = 50^{\circ}$ and $\angle B = 65^{\circ}$,then $\angle ACD = \dots$ (in $^{\circ}$)

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