Question

The shadow of a tower standing on the level ground is found to be $40$ m shorter when the sun's altitude changes from ${30}^o$ to ${60}^o$, find the height of tower.

Answer 1

Given, tower be $AB$

when sun's altitude is ${60}^0$

$\angle ACB={60}^0$

& length of shadow $=BC$

when sun's altitude is ${30}^0$

$\angle ADB={30}^0$

& length of shadow $=DB$

since the tower is vertical to ground

$\therefore \angle ABC={90}^0$

$△ABC$

$tanC=\frac{AB}{CB}$

$\Rightarrow {\tan 60}^0=\frac{AB}{CB}$

$\Rightarrow CB=\frac{AB}{\sqrt{3}}⟶\left(1\right)$

$\{\because {tan60}^0=\sqrt{3}\}$

in $△ABD$

$tanD=\frac{AB}{DB}$

$\Rightarrow {\tan 30}^0=\frac{AB}{DB}$

$\Rightarrow DB=\sqrt{3}AB\{\because {\tan 30}^0=\frac{1}{\sqrt{3}}\}$

$\Rightarrow DC+CB=\sqrt{3}AB$

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

$\Rightarrow CB=\sqrt{3}AB-40⟶\left(2\right)$

from $\left(1\right)$ & $\left(2\right)$

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

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

$\Rightarrow 2AB=40\sqrt{3}$

$AB=20\sqrt{3}=$ Heigth of tower