MCQ
In a $\triangle A B C, B C=A B$ and $\angle B=80^{\circ}$, then $\angle A=$
- A$80^{\circ}$
- B$50^{\circ}$
- C$40^{\circ}$
- D$100^{\circ}$
✓
Answer
B. $50^{\circ}$
We have,
$B C=A B$ and $\angle B=80^{\circ}$$
\begin{array}{ll}
\therefore & \angle A=\angle C \text { and } \angle B=80^{\circ} \\
\text { Now } & \angle A+\angle B+\angle C=180^{\circ} \Rightarrow \angle A+80^{\circ}+\angle A=180^{\circ} \Rightarrow 2 \angle A=100 \Rightarrow \angle A=50^{\circ}
\end{array}
$
We have,
$B C=A B$ and $\angle B=80^{\circ}$$
\begin{array}{ll}
\therefore & \angle A=\angle C \text { and } \angle B=80^{\circ} \\
\text { Now } & \angle A+\angle B+\angle C=180^{\circ} \Rightarrow \angle A+80^{\circ}+\angle A=180^{\circ} \Rightarrow 2 \angle A=100 \Rightarrow \angle A=50^{\circ}
\end{array}
$
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