Question

In the adjacent figure, it is given that AB = AC, $\angle BAC={36}^o,\angle ADB={45}^o$ and $\angle AEC={40}^o$. Find $\angle ACB$.

Answer 1

In $△ABC$

$AB=AC$

So $△ABC$ is isoceles.

$\Rightarrow \angle ABC=\angle ACB$

$\angle BAC+\angle ABC+\angle ACB={180}^{\circ }$ (Angle sum property of triangle)

${36}^{\circ }+\angle ACB+\angle ACB={180}^{\circ }2\angle ACB={180}^{\circ }-{36}^{\circ }2\angle ACB={144}^{\circ }\angle ACB=\frac{{144}^{\circ }}{2}={72}^{\circ }$