Question

Mark (✓) against the correct answer
In the given figure, side BC of $\Delta ABC$ is produced to D such that $\angle ABC={70}^{\circ }$ and $\angle ACD={120}^{\circ }$. Then, $\angle BAC=$?

(a) ${60}^{\circ }$
(b) ${50}^{\circ }$
(c) ${70}^{\circ }$
(d) ${35}^{\circ }$

Answer 1

The correct option is (b): ${50}^{\circ }$

In triangle ABC, $\angle ACD$ is the exterior angle.

$\angle ABC+\angle BAC=\angle ACD$ [ Exterior angle is equal to opposite interior angles] ]

$\Rightarrow {70}^{\circ }+\angle BAC={120}^{\circ }$

$\Rightarrow \angle BAC={120}^{\circ }-{70}^{\circ }={50}^{\circ }$