Question

The shadow of a tower standing on a level ground is found to be $40m$ longer when the sun’s altitude is ${30}^{\circ }$ than when it is ${60}^{\circ }$. Find the height of the tower.

Answer 1

Let AB be the tower, BC be the shadow at ${60}^{\circ },$ BD be the shadow at ${30}^{\circ }$.

The shadow of a tower standing on a level ground is found to be $40m$ longer when the sun’s altitude is ${30}^{\circ }$ than when it is ${60}^{\circ }.$

i.e. $CD=40m$

Let $BC=xm$

In triangle ABC, we have

$\tan {60}^{\circ }=\frac{AB}{BC}$

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

$\Rightarrow x=\frac{AB}{\sqrt{3}}\dots \dots \left(i\right)$

In triangle ABD, we have

$\tan {30}^{\circ }=\frac{AB}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{x+40}$

$\Rightarrow x+40=\sqrt{3}AB$

$\Rightarrow x=\sqrt{3}AB-40\dots \dots \left(ii\right)$

From eq.(i) and (ii), we get

$\frac{AB}{\sqrt{3}}=\sqrt{3}AB-40$

$\Rightarrow AB=3AB-40\sqrt{3}$

$\Rightarrow AB=\frac{40\sqrt{3}}{2}=20\sqrt{3}m$

Hence, height of the tower is $20\sqrt{3}m$.