Question

In figure, $\angle$ACD is an exterior angle of $\Delta$ABC. $\angle B={40}^o$, $\angle A={70}^o$. Find the measure of $\angle$ACD.

Answer 1

We are given $m\angle B={40}^o$

$m\angle A={70}^o$

Now, $\overrightarrow{BD}$ is a straight ray

so, $m\angle ACB+m\angle ACD={180}^o\to \left(1\right)$

In $△ABC,m\angle A+m\angle B+m\angle ACB={180}^o\to \left(2\right)$

From $\left(1\right)$ & $\left(2\right)$

$m\angle A+m\angle B=m\angle ACD$

so, $m\angle ACD={40}^o+{70}^o$

$m\angle ACD={110}^o$