Question

In the figure, the side $BCof△ABC$ is extended to $D.\angle BAC=\_\_\_\_\_\_\_\_\_.$

Answer 1

Since the side $BC$ is extended to$D.$

So, $\angle ABC$ is an exterior angle.

and we know that exterior angle is the sum of two interior opposite angles.

Given $$\begin{gathered} \angle ABC=48° \\ \angle ACD=120° \end{gathered}$$

hence,

$$\begin{gathered} \angle ABC+\angle BAC=\angle ACD \\ \Rightarrow 48°+\angle BAC=120° \\ \Rightarrow \angle BAC=120°-48° \\ \Rightarrow \angle BAC=72° \end{gathered}$$

Hence, $\angle BAC=72°$