Question

$ABCD$ is a rectangle such that the length $BC$ is twice the width $AB.$ Pick a point $P$ on side $BC$ such that the lengths of $AP$ and $BC$ are equal. Then $\angle CPD$ is

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${75}^{\circ }$
• D. none of the above

Answer 1

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

Let $AB=a$
Then $BC=2a$

In right triangle $ABP,$
$A{B}^2+B{P}^2=A{P}^2$
$\therefore BP=\sqrt{3}a$

$PC=BC-BP=(2-\sqrt{3})a$

Now, in right triangle $CPD,$
$\tan \left(\angle CPD\right)=\frac{CD}{PC}$
$=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$
$\therefore \angle CPD={75}^{\circ }$