Question

From a point on the ground the angles of elevation of the bottom and top of a communication tower fixed on the top of a 20-m-high building are 45° and 60° respectively. Find the height of the tower. $[\mathrm{Take}\sqrt{3}=1.732]$ [CBSE 2017]

Answer 1

Let BC be the 20 m high building and AB be the communication tower of height h fixed on top of the building. Let D be a point on ground such that CD = x m and angles of elevation made from this point to top and bottom of tower are $45°$ and $60°$.
In $△\mathrm{BCD}:\tan 45°=\frac{\mathrm{BC}}{\mathrm{CD}}=\frac{20}{x}$
$\Rightarrow 1=\frac{20}{x}\Rightarrow x=20\mathrm{m}.$

Also, in $△\mathrm{ACD}:\tan 60°=\frac{\mathrm{AC}}{\mathrm{CD}}=\frac{20+h}{x}$
$$\begin{gathered} \Rightarrow \sqrt{3}=\frac{20+h}{x}\Rightarrow \sqrt{3}=\frac{20+h}{20} \\ \Rightarrow 20+h=20\sqrt{3}\Rightarrow h=20\sqrt{3}-20 \\ \Rightarrow h=20\left(\sqrt{3}-1\right)=20\left(1.73-1\right)=20\times 0.73 \\ \Rightarrow h=14.64\mathrm{m}. \end{gathered}$$