Question

The shadow of a tower standing on a level plane is found to be $50\mathrm{m}$longer when Sun’s elevation is $30°$ than when it is$60°.$ Find the height of the tower.

Answer 1

Step 1: Write the value of $x$ in $∆SRQ$

$\tan \theta =\frac{perpendicular}{base}$

$\Rightarrow$$\tan 60°=\frac{\mathrm{h}}{x}$

$\Rightarrow$ $\sqrt{3}=\frac{\mathrm{h}}{x}$

$\Rightarrow$ $x=\frac{\mathrm{h}}{\sqrt{3}}$

Step 2: Write the value of $x$ in $∆SPQ$

$\tan 30°=\frac{\mathrm{h}}{50+x}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{50+x}$

$\Rightarrow$$50+\mathrm{x}=\mathrm{h}\sqrt{3}$

$\Rightarrow$ $\mathrm{x}=\mathrm{h}\sqrt{3}-50$

Step 3: Compare the value of $x$ in both the triangles

On comparing, we get: $\frac{h}{\sqrt{3}}=\mathrm{h}\sqrt{3}-50$

$\Rightarrow$ $h=3\mathrm{h}-50\sqrt{3}$

$\Rightarrow$ $50\sqrt{3}=3\mathrm{h}-h$

$\Rightarrow$ $50\sqrt{3}=2\mathrm{h}$

$\Rightarrow$ $25\sqrt{3}=\mathrm{h}$

Hence, the height of the tower is $25\sqrt{3}m$.