Question

In the figure, $ \mathrm{BC}=\mathrm{CA}$ and $ \angle \mathrm{A}=40°.$ Then, $ \angle \mathrm{ACD} $is equal to

• A. $40°$
• B. $80°$
• C. $120°$
• D. $60°$

Answer 1

The correct option is B

$80°$

The explanation for the correct answer.

Find the required angle:

Given: $\mathrm{BC}=\mathrm{CA}$ and $\angle \mathrm{A}=40.$

$\mathrm{In}∆\mathrm{ABC},\mathrm{AC}=\mathrm{BC}$

$\therefore \angle \mathrm{A}=\angle \mathrm{B}=40°$ ($∆\mathrm{ABC}$ is an isosceles triangle)

Since, $\angle \mathrm{ACD}$ is an exterior triangle.

$\angle \mathrm{ACD}=\angle \mathrm{A}+\angle \mathrm{B}$ (Exterior angle is equal to the sum of opposite interior angles)

$$\begin{gathered} \angle \mathrm{ACD}=40°+40° \\ \angle \mathrm{ACD}=80° \end{gathered}$$

Hence option $B$ is the correct answer.