The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(20\sqrt{3}\) m
The Sun’s altitude is the angle of elevation of the top of the tower seen from the tip of its shadow, so \(\tan(\text{altitude})=\frac{\text{height}}{\text{shadow}}\). The lower the Sun, the longer the shadow.
Tower AB \(=h\); shadow BC \(=x\) at \(60^\circ\), shadow BD \(=x+40\) at \(30^\circ\)
Sun at \(60^\circ\), shadow \(x\): \(\tan60^\circ=\frac{h}{x}\), so \(\sqrt{3}=\frac{h}{x}\) and \(x=\frac{h}{\sqrt{3}}\) (1)
Sun at \(30^\circ\), shadow \(x+40\): \(\tan30^\circ=\frac{h}{x+40}\), so \(\frac{1}{\sqrt{3}}=\frac{h}{x+40}\) and \(x+40=h\sqrt{3}\) (2)
Subtracting (1) from (2), \(x\) cancels: \(\displaystyle 40=h\sqrt{3}-\frac{h}{\sqrt{3}}=\frac{3h-h}{\sqrt{3}}=\frac{2h}{\sqrt{3}}\)
Height: \(h=\frac{40\sqrt{3}}{2}=20\sqrt{3}\,\mathrm{m}\)