Question

Question 16
The angle of elevation of the top of a vertical tower from a point on the ground is ${60}^{\circ }$. From another point 10m vertically above the first, its angle of elevation is ${45}^{\circ }$. Find the height of the tower.

Answer 1

Let the height the vertical tower, OT = H

And OP = AB = x m

Given that, AP = 10 m

And $\angle TPO={60}^{\circ },\angle TAB={45}^{\circ }$

Now, in $\Delta TPO$, $tan{60}^{\circ }=\frac{OT}{OP}=\frac{H}{x}$

$\Rightarrow \sqrt{3}=\frac{H}{x}$

$\Rightarrow x=\frac{H}{\sqrt{3}}$ ...(i)

And in $\Delta TAB$, $tan{45}^{\circ }=\frac{TB}{AB}=\frac{H-10}{x}$

$\Rightarrow 1=\frac{H-10}{x}\Rightarrow x=H-10$

$\Rightarrow \frac{H}{\sqrt{3}}=H-10$ [from Eq. (i)]

$\Rightarrow H-\frac{H}{\sqrt{3}}=10\Rightarrow H(1-\frac{1}{\sqrt{3}})=10$

$\Rightarrow H\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)=10$

$\therefore H=\frac{10\sqrt{3}}{\sqrt{3}-1}.\frac{\sqrt{3}+1}{\sqrt{3}+1}$ [By rationalization]

$=\frac{10\sqrt{3}(\sqrt{3}+1)}{3-1}=\frac{10\sqrt{3}(\sqrt{3}+1)}{2}$

$\Rightarrow =5\sqrt{3}(\sqrt{3}+1)=5(\sqrt{3}+3)m$