On the level ground, the angle of elevation of a tower is 30°. On moving 20 m nearer, the angle of elevation is 60°. The height of the tower is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(10\sqrt{3}\) m
Let AB = \(h\) be the tower, D the nearer point and DB = \(x\). In each right triangle, tan of the angle of elevation = height ÷ distance from the foot.
From D: \(\tan60^{\circ}=\displaystyle\frac{h}{x}\), so \(h=\sqrt{3}\,x\) (1)
From C: \(\tan30^{\circ}=\displaystyle\frac{h}{x+20}\), so \(\sqrt{3}\,h=x+20\) (2)
Putting (1) in (2): \(3x=x+20\), so \(x=10\) m
\(h=10\sqrt{3}\) m