Question

In the figure shows a square $ABCD$ and an equilateral triangle $ABP$.
calculate $\angle BPC$

• A. ${45}^{\circ }$
• B. ${75}^{\circ }$
• C. ${55}^{\circ }$
• D. ${65}^{\circ }$

Answer 1

The correct option is B ${75}^{\circ }$

In the given figure,$△APB$ is an equilateral triangle.

$\therefore$all its angles are ${60}^{\circ }$

Again in the $△ADB,$

$\angle ABD={45}^{\circ }$

$\angle AOB={180}^{\circ }-{60}^{\circ }-{45}^{\circ }={75}^{\circ }$

Again,$△BPC$

$\Rightarrow \angle BPC={75}^{\circ }$ since $BP=CB$