Question

Height of tower AB is 30 m where B is foot of tower. Angle of elevation from a point C on level ground to top of tower is 60° and angle of elevation of A from a point D x m above C is 15° then find the area of quadrilateral ABCD.

• A. 300($\sqrt{3}$-1)
• B. 600($\sqrt{3}$-1)
• C. 150($\sqrt{3}$-1)
• D. 100($\sqrt{3}$-1)

Answer 1

Answer: (B)

600($\sqrt{3}$-1)

Answer 2

tan 60° = $\frac{30}{y}$ = $\sqrt{3}$
$\Rightarrow$ y = 10$\sqrt{3}$
tan 15° = $\frac{30-x}{y}$
(2 - $\sqrt{3}$)10$\sqrt{3}$ = 30-x
x = 30-20$\sqrt{3}$ + 30
x = 60-20$\sqrt{3}$
Area of ABCD = xy =(60-2$\sqrt{3}$).10$\sqrt{3}$
= 600($\sqrt{3}$-1)