Question

A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR = 15 m. If from a point A on the ground the angle of elevation of R is 60° and the part PR of the tower subtends an angle of 15° at A, then the height of the tower is :

• A. 5(2$\sqrt3$+3)m
• B. 5($\sqrt3$+3)m
• C. 10($\sqrt3$+1)m
• D. 10(2$\sqrt3$+1)m

Answer 1

Answer: (A)

5(2$\sqrt3$+3)m

Answer 2

From △APQ
$\frac{x+15}{y}$=tan⁡75º⋯(i)
From △RQA
$\frac{15}{y}$=tan⁡60º⋯(ii)
From (i) and (ii)
$\frac{x+15}{15}$=tan⁡75ºtan⁡60º=tan⁡(45º+30º)tan⁡60º=3+1(3−1)⋅3
On simplification,
x=10$\sqrt3$ m
Hence the height of the tower
=(15+10$\sqrt3$) m
=5(2$\sqrt3$+3) m