Question

The shadow of a tower, when the angle of elevation of the sun is $30^\circ$, is $50$ m longer than when the angle of elevation was $60^\circ$ on the plane ground. Find the height of the tower.

Hint:

For sun–shadow problems, write shadow $=h\cot\alpha$ and use the given difference to get a single linear equation in $h$.

Answer 1

Let the height of the tower be $h$.
Shadow length at altitude $\alpha$ is $h\cot\alpha$.
At $30^\circ$: shadow $=h\cot30^\circ=h\sqrt{3}$.
At $60^\circ$: shadow $=h\cot60^\circ=\dfrac{h}{\sqrt{3}}$.
Given: $h\sqrt{3}-\dfrac{h}{\sqrt{3}}=50 $\Rightarrow$ h\left(\dfrac{3-1}{\sqrt{3}}\right)=50 $\Rightarrow$ h\cdot\dfrac{2}{\sqrt{3}}=50 $\Rightarrow$ h=25\sqrt{3}\ \text{m}.
[1mm] \boxed{h=25\sqrt{3}\ \text{m}\ \ (\approx 43.3\ \text{m})}