Question

Question 19

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

a) ${40}^{\circ }$
b) ${80}^{\circ }$
c) ${120}^{\circ }$
d) ${60}^{\circ }$

Answer 1

Given, BC = CA
$\therefore \angle B=\angle A={40}^{\circ }$ [$\because$ opposite angles of two equal sides are equal]
As we know, the measure of any exterior angle of a triangle is equal to the sum of the measure of its two interior opposite angles
So, $\angle ACD=\angle A+\angle B={40}^{\circ }+{40}^{\circ }$
$\Rightarrow \angle ACD={80}^{\circ }$