Question

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

Answer 1

We have $AB=AC$ and $\angle ACD={105}^{\circ }$

Since,$\angle BCD={180}^{\circ }=$Straight angle.

$\Rightarrow \angle BCA+\angle ACD={180}^{\circ }$

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

$\Rightarrow \angle BCA={180}^{\circ }-{105}^{\circ }={75}^{\circ }$ ......$\left(1\right)$

And also, $△ABC$ is an isosceles triangle ($AB=AC$)

$\Rightarrow \angle ABC=\angle ACB$ since angles opposite to equal sides are equal.

From $\left(1\right)$, we have

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

And also,

Sum of interior angles of a triangle$={180}^{\circ }$

$\Rightarrow \angle ABC+\angle BCA+\angle CAB={180}^{\circ }$

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

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

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