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Question

(D) In $\Delta ABC$,the exterior angle at vertex $B$ is $\angle ABD = 140^{\circ}$. Since the sum of an interior angle and its adjacent exterior angle

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) In $\Delta ABC$,the exterior angle at vertex $B$ is $\angle ABD = 140^{\circ}$. Since the sum of an interior angle and its adjacent exterior angle is $180^{\circ}$,we have $\angle ABC = 180^{\circ} - 140^{\circ} = 40^{\circ}$.Similarly,the exterior angle at vertex $C$ is $\angle ACE = 80^{\circ}$. Thus,the interior angle $\angle ACB = 180^{\circ} - 80^{\circ} = 100^{\circ}$.In $\Delta ABC$,the sum of all interior angles is $180^{\circ}$. Therefore,$\angle A + \angle ABC + \angle ACB = 180^{\circ}$.Substituting the values,$\angle A + 40^{\circ} + 100^{\circ} = 180^{\circ}$.$\angle A + 140^{\circ} = 180^{\circ}$.$\angle A = 180^{\circ} - 140^{\circ} = 40^{\circ}$.

$\angle ABD$ and $\angle ACE$ are exterior angles of $\Delta ABC$. If $\angle ABD = 140^{\circ}$ and $\angle ACE = 80^{\circ}$,then find $\angle A$. (in $^{\circ}$)

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