The shadow of a tower standing on level ground is found to be $40$ m longer when the sun's altitude (angle of elevation) is $30^\circ$ than when it is $60^\circ$. Find the height of the tower.
Step-by-Step Solution
Key Concept: Set up two equations for the two shadow lengths using the same height, subtract to eliminate the unknown shadow length at 60°.
Let the height be $h$ and the shorter shadow (at $60^\circ$) be $s$. At $60^\circ$: $\tan60^\circ=\dfrac{h}{s}\Rightarrow s=\dfrac{h}{\sqrt3}$. [1.0 Mark]
At $30^\circ$, the shadow is $s+40$: $\tan30^\circ=\dfrac{h}{s+40}\Rightarrow s+40=h\sqrt3$. [1.0 Mark]
Substituting: $\dfrac{h}{\sqrt3}+40=h\sqrt3\Rightarrow 40=h\sqrt3-\dfrac{h}{\sqrt3}=h\left(\sqrt3-\dfrac{1}{\sqrt3}\right)=h\times\dfrac{2}{\sqrt3}\Rightarrow h=20\sqrt3$ m. [1.0 Mark]
Correct Answer: