MCQ
In Fig , $\angle A C D=120^{\circ}$ and $\angle A B C=40^{\circ}$, then $\angle B A C=$
- A$80^{\circ}$
- B$60^{\circ}$
- C$50^{\circ}$
- D$40^{\circ}$
✓
Answer
A. $80^{\circ}$
In $\triangle A B C$, side $B C$ is produced to $D$. Using exterior angle property, we obtain$
\begin{array}{ll}
& \angle A C D=\angle A B C+\angle B A C \\
\Rightarrow \quad & 120^{\circ}=40^{\circ}+\angle B A C \Rightarrow \angle B A C=80^{\circ}
\end{array}
$
In $\triangle A B C$, side $B C$ is produced to $D$. Using exterior angle property, we obtain$
\begin{array}{ll}
& \angle A C D=\angle A B C+\angle B A C \\
\Rightarrow \quad & 120^{\circ}=40^{\circ}+\angle B A C \Rightarrow \angle B A C=80^{\circ}
\end{array}
$
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