Question

ABCD is a square. P is a point in its interior such that $\Delta$PAB is equilateral. Find $\tan \angle$PCB is ...........

• A. $\frac{1}{2+\sqrt{3}}$
• B. $\frac{1}{2-\sqrt{3}}$
• C. $\frac{1}{4-\sqrt{3}}$
• D. $\frac{1}{4+\sqrt{3}}$

Answer 1

Answer: (B)

$\frac{1}{2-\sqrt{3}}$

Given that $△APB$ is an equilateral triangle.

$\therefore LPBA={60}^0\therefore LPBC=90-60={30}^0$

(Complementary angle)

In$\Delta PCBPB=BC⟹(LBPC=LPCB)⟹LPBC+LBPC+LPCB={180}^0$

$⟹2LPCB=180-30⟹LPCB={75}^0\therefore \tan \left(LPCB\right)=\tan 75=\frac{1}{2-\sqrt{3}}$