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

The Correct Option is B

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)