In ∆ABC, AB = AC and ∠B = 50°.


In ABCAB AC and ∠B = 50°. Then, ∠A = ?
(a) 40°
(b) 50°
(c) 80°
(d) 130°


In $\triangle A B C$, we have:

$A B=A C$

$\angle B=50^{\circ}$

Since $A B C$ is an isosceles triangle, we have:

$\angle C=\angle B$

$\angle C=50^{\circ}$

In triangle $\mathrm{ABC}$, we have:

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow \angle A+50+50=180^{\circ}$

$\Rightarrow \angle A=180^{\circ}-100^{\circ}$

$\therefore \angle A=80^{\circ}$

Hence, the correct answer is option (c).

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