In triangle ABC,BC = AB and angle B = 80^{cir | vedclass

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(A) In $	riangle ABC$,we are given that $BC = AB$.Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equa

Vedclass · Accepted Answer

$50$

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(A) In $	riangle ABC$,we are given that $BC = AB$.Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal.Therefore,$\angle A = \angle C$.We know that the sum of all interior angles of a triangle is $180^{\circ}$.So,$\angle A + \angle B + \angle C = 180^{\circ}$.Substituting the given values: $\angle A + 80^{\circ} + \angle A = 180^{\circ}$.$2 \angle A + 80^{\circ} = 180^{\circ}$.$2 \angle A = 180^{\circ} - 80^{\circ}$.$2 \angle A = 100^{\circ}$.$\angle A = 100^{\circ} / 2 = 50^{\circ}$.

In $\triangle ABC$,$BC = AB$ and $\angle B = 80^{\circ}$. Then $\angle A$ is equal to: (in $^{\circ}$)

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