Question

In the figure, AB = AC and $\angle ACD={105}^{\circ },find\angle BAC.$

Answer 1

In $\Delta ABC,AB=AC$

$\therefore \angle B=\angle C$ (Angles opposite to equal sides)

But $\angle ACB+\angle ACD={180}^{\circ }$ (Linear pair)

$\Rightarrow \angle ACB+{105}^{\circ }={180}^{\circ }$

$\Rightarrow ACD={180}^{\circ }-{105}^{\circ }={75}^{\circ }$

$\therefore \angle ABC=\angle ACB={75}^{\circ }$

But $\angle A+\angle B+\angle C={180}^{\circ }$

(Sum of angles of a triangle)

$\Rightarrow \angle A+{75}^{\circ }+{75}^{\circ }={180}^{\circ }$

$\Rightarrow \angle A+{150}^{\circ }={180}^{\circ }$

$\Rightarrow \angle A={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

$\therefore \angle BAC={30}^{\circ }$