Question

In the figure, $△ABC$ is equilateral and $△PBC$ is isosceles.If $\angle PBA={20}^{\circ };$Find $\left(i\right)\angle PBC$ and $\left(ii\right)\angle BPC$

Answer 1

In the figure $△ABC$ is an equilateral triangle and $△PBC$ is an isosceles triangle and $\angle PBA={20}^{\circ }$

$\because △ABC$ is an equilateral triangle.

$\therefore \angle ABC={60}^{\circ }$

$\Rightarrow \angle PBA+\angle PBC={60}^{\circ }$

$\Rightarrow {20}^{\circ }+\angle PBC={60}^{\circ }$

$\Rightarrow \angle PBC={60}^{\circ }-{20}^{\circ }={40}^{\circ }$

$\because \angle PBC$ is an isosceles triangle.

$\therefore \angle PBC=\angle PCB={40}^{\circ }$

Now, in $△BPC,$

$\angle PBC+\angle PCB+\angle BPC={180}^{\circ }$

$\Rightarrow {80}^{\circ }+\angle BPC={180}^{\circ }$

$\Rightarrow \angle BPC={180}^{\circ }-{80}^{\circ }={100}^{\circ }$