The angle of elevation of the top of a tower from a point on the horizontal is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is: (Take \(\sqrt{3}=1.73\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 27.3 m
The new angle is \(30^\circ+15^\circ=45^\circ\). Let the tower TB have height \(h\), and let Q, the second point, be \(x\) m from its foot. Each angle of elevation gives a right triangle, and tan of the angle = height ÷ distance from the foot.
At Q: \(\tan45^\circ=\frac{h}{x}=1\), so \(x=h\)
At P, 20 m farther: \(\displaystyle \tan30^\circ=\frac{h}{x+20}=\frac{1}{\sqrt3}\), so \(h\sqrt3=h+20\)
\(h(\sqrt3-1)=20\)
\(\displaystyle h=\frac{20}{\sqrt3-1}=\frac{20(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\frac{20(\sqrt3+1)}{2}=10(\sqrt3+1)\)
\(h=10\times2.73=27.3\) m