Question

The shadow of a tower on a level plane is found to be $60$ metres longer when the sun's altitude is ${30}^0$ than that when it is ${45}^0$. The height of the tower in metres is

• A. $30(\sqrt{3}+1)$
• B. $30(\sqrt{3}-1)$
• C. $30(3+\sqrt{3})$
• D. $30(3-\sqrt{3})$

Answer 1

Answer: (A)

$30(\sqrt{3}+1)$

Let height of tower be $hm$.

And length of shadow when sun's altitude is at ${30}^0$.

$\tan {45}^0=\frac{h}{L_1}$

$\Rightarrow L_1=h$

$\tan {30}^0=\frac{h}{L_1+60}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{L_1}{L_1+60}$

$\Rightarrow \sqrt{3}=1+\frac{60}{L_1}$

$\Rightarrow L_1=\frac{60}{(\sqrt{3}-1)}$

$\Rightarrow L_1=30(\sqrt{3}+1)m$