MCQ
In Figure. $BC = CA$ and $\angle\text{A} = 40$. Then, $\angle\text{ACD}$ is equal to:
- A$40^\circ$
- ✓$80^\circ$
- C$120^\circ$
- D60°
✓
Answer
Correct option: B.
$80^\circ$
$\text{Given}, \ \text{BC}=\text{CA},$
$\therefore\angle\text{B}=\angle\text{A}=40^{\circ}$ $[\because$ opposite angles of two equal sides are equal$]$
As we know, the measure of any exterior angles of a triangle is equal to the sum of the measure of its two interior opposite angles.
So, $\angle\text{ACD}=\angle\text{A}+\angle\text{B}=40^{\circ}+40^{\circ}$
$\angle\text{ACD}=80^{\circ}$.
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