Question

From the top of a building $16m$ high, the angular elevation of the top of a hill is ${60}^o$ and the angular depression of the foot of the hill is ${30}^o$. Find the height of the hill.

Answer 1

Let $x$ be the total height of the hill. Therefore, the height of the hill above the building is $(x-16)$ and let $y$ be the perpendicular distance from the top of the building to the hill.

We know that $\tan \theta =\frac{Oppositeside}{Adjacentside}=\frac{BC}{AB}$

In $△ABC,$ $\angle B={90}^0$ and $\angle A={30}^0$.

Here, $\theta ={30}^0$, $BC=16$ m and $AB=y$ m, therefore,

$\tan \theta =\frac{BC}{AB}\Rightarrow \tan {30}^0=\frac{16}{y}\Rightarrow \frac{1}{\sqrt{3}}=\frac{16}{y}(\because \tan {30}^0=\frac{1}{\sqrt{3}})\Rightarrow y=16\sqrt{3}...........\left(1\right)$

Also, in $△ABD,$ $\angle B={90}^0$ and $\angle A={60}^0$.

Here, $\theta ={60}^0$, $BD=(x-16)$ m and $AB=y$ m, therefore,

$\tan \theta =\frac{BD}{AB}\Rightarrow \tan {60}^0=\frac{x-16}{y}\Rightarrow \sqrt{3}=\frac{x-16}{y}(\because \tan {60}^0=\sqrt{3})\Rightarrow \sqrt{3}y=x-16...........\left(2\right)$

Substitute the value of equation 1 in equation 2, we get

$\sqrt{3}\times 16\sqrt{3}=x-16\Rightarrow 3\times 16=x-16\Rightarrow 48=x-16\Rightarrow x=48+16=64$

Hence, the height of the hill is $64$ m.