Question

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

Answer 1

Let, $AB$ be the height of the tower $=h$

From $\Delta ABD$

$tan{60}^o=\frac{AB}{BD}$

$\Rightarrow \sqrt{3}=\frac{h}{BD}$

$\Rightarrow h=BD\sqrt{3}$

$\Rightarrow BD=\frac{h}{\sqrt{3}}..........\left(1\right)$

From $\Delta AEC$

$tan45=\frac{AE}{EC}$

$\Rightarrow 1=\frac{AE}{EC}$

$\Rightarrow AE=EC$

$\Rightarrow AE=BD$

$\Rightarrow AE=\frac{h}{\sqrt{3}}..........\left(2\right)$

We can write, $AE=AB-BE=h-10.........\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$

$h-10=\frac{h}{\sqrt{3}}$

$\Rightarrow h\sqrt{3}-10\sqrt{3}=h$

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

$\Rightarrow h=\frac{10\sqrt{3}}{\sqrt{3}-1}$

$\Rightarrow h=\frac{10\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

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

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

$\Rightarrow h=15+5\sqrt{3}$

$\Rightarrow h=15+5\times 1.732$

$\Rightarrow h=23.66$

Hence, the height of the tower is $23.66m$