In ΔABC,if BC = AB and ∠B = 80°,

(c)

Given, $\triangle A B C$ such that $B C=A B$ and $\angle B=80^{\circ}$

$\ln \triangle A B C$, $A B=B C$

$\Rightarrow \quad \angle C=\angle A$ $\ldots(1)$

[angles opposite to equal sides are equal]

We know that, the sum of all the angles of a triangle is $180^{\circ}$.

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

$\Rightarrow \quad \angle A+80^{\circ}+\angle A=180^{\circ}$

$\Rightarrow \quad 2 \angle A=180^{\circ}-80^{\circ}=100^{\circ}$

$\Rightarrow \quad \angle A=\frac{100^{\circ}}{2}$

$\Rightarrow \quad \angle A=50^{\circ}$