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(D) In $\Delta ABC$,the interior angles are $\angle ABC$,$\angle ACB$,and $\angle BAC$ (or $\angle A$).Since $\angle ABD$ is an exterior angle at vert

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$80$

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(D) In $\Delta ABC$,the interior angles are $\angle ABC$,$\angle ACB$,and $\angle BAC$ (or $\angle A$).Since $\angle ABD$ is an exterior angle at vertex $B$,we have $\angle ABC + \angle ABD = 180^{\circ}$ (linear pair).$\angle ABC = 180^{\circ} - 110^{\circ} = 70^{\circ}$.Since $\angle ACE$ is an exterior angle at vertex $C$,we have $\angle ACB + \angle ACE = 180^{\circ}$ (linear pair).$\angle ACB = 180^{\circ} - 150^{\circ} = 30^{\circ}$.In $\Delta ABC$,the sum of interior angles is $180^{\circ}$,so $\angle A + \angle ABC + \angle ACB = 180^{\circ}$.$\angle A + 70^{\circ} + 30^{\circ} = 180^{\circ}$.$\angle A + 100^{\circ} = 180^{\circ}$.$\angle A = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

$\angle ABD$ and $\angle ACE$ are exterior angles of $\Delta ABC$. If $\angle ABD = 110^{\circ}$ and $\angle ACE = 150^{\circ}$,then $\angle A$ is: (in $^{\circ}$)

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