The angle of elevation of the top of a tower from a certain point is 60°. If the observer moves 10 m away from the tower, the angle of elevation of the top of the tower decreases by 15°. The height of the tower is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 23.65 m
The tangent of the angle of elevation is the height of the tower divided by the distance of the observer from its foot. Moving 10 m away lowers the angle by 15°, from 60° to 45°.
The two angles of elevation (tangent = height ÷ distance from the foot):
From C: \(\displaystyle \tan60^\circ=\frac{h}{x}\)
So \(h=\sqrt{3}\,x\) (1)
From D: \(\displaystyle \tan45^\circ=\frac{h}{x+10}\)
So \(h=x+10\) (2)
Finding \(x\) (equating (1) and (2), \(h\) drops out):
\(\sqrt{3}\,x=x+10\)
\(\displaystyle x=\frac{10}{\sqrt{3}-1}\)
\(\displaystyle x=\frac{10(\sqrt{3}+1)}{3-1}=5(\sqrt{3}+1)\)
\(x=5\times2.73=13.65\) m
The height (from (2)):
\(h=x+10=23.65\) m