In Delta ABC,angle B = 50^{circ} and angle C | vedclass

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Question

(D) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$\angle A + 50^{\circ} + 85^{\circ

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(D) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$\angle A + 50^{\circ} + 85^{\circ} = 180^{\circ}$.
$\angle A + 135^{\circ} = 180^{\circ}$.
$\angle A = 180^{\circ} - 135^{\circ} = 45^{\circ}$.
We know that in a triangle,the side opposite to the larger angle is longer.
Here,$\angle C = 85^{\circ}$ and $\angle B = 50^{\circ}$.
Since $\angle C > \angle B$,the side opposite to $\angle C$ $(AB)$ is greater than the side opposite to $\angle B$ $(AC)$.
Thus,$AB > AC$.

In $\Delta ABC$,$\angle B = 50^{\circ}$ and $\angle C = 85^{\circ}$,then $AB$ $\dots$ $AC$.

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