Question

A tall building casts a shadow of $300m$ long when the sun's altitude (elevation) is ${30}^o$. Find the height of the tower.

Answer 1

Let $△ABC$ be a right angled triangle where $\angle B={90}^0$ and $\angle C={30}^0$ as shown in the above figure.

Let $x$ be the height of the tower.

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

Here, $\theta ={30}^0$, $BC=300$ m and $AB=x$ m, therefore,

$\tan \theta =\frac{AB}{BC}\Rightarrow \tan {30}^0=\frac{x}{300}\Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{300}(\because \tan {30}^0=\frac{1}{\sqrt{3}})\Rightarrow \sqrt{3}x=300\Rightarrow x=\frac{300}{\sqrt{3}}\Rightarrow x=\frac{300\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\frac{300\times \sqrt{3}}{3}=100\sqrt{3}$

Hence, the height of the tower is $100\sqrt{3}$ m.