Question

The angle of elevation of the top of a tower from a point A due south of the tower is ${30}^{\circ }$ and from B due east of the tower is ${60}^{\circ }.$ If AB = 60 km then find the height of the tower.

• A. 35 km
• B. 36.2km
• C. 32.8km
• D. 39km

Answer 1

The correct option is C 32.8km

Let CD be a tower of height ‘h’ m. A and B are two stations due south and east of tower CD such that AB = 60 km. The angle of elevation of top of tower CD from A and B are ${30}^{\circ }$ and ${60}^{\circ }$ respectively. Let AD = x km and BD = ‘y’ km.
In $\Delta ACD,\tan {30}^{\circ }=\frac{CD}{AD}=\frac{h}{x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}\Rightarrow x=\sqrt{3}h$ ....(1)
In $\Delta BCD,\tan {60}^{\circ }=\frac{CD}{BD}=\frac{h}{y}$
$\Rightarrow \sqrt{3}=\frac{h}{y}\Rightarrow y=\frac{h}{\sqrt{3}}$ ....(2)
In ΔABD,
$A{D}^2+B{D}^2=A{B}^2$ (by pythagoras theorem)
$\Rightarrow x^2+y^2=(60{)}^2$
$\Rightarrow {\left(\sqrt{3}h\right)}^2+{\left(\frac{h}{\sqrt{3}}\right)}^2={\left(60\right)}^2$
$\Rightarrow 3h^2+\frac{h^2}{3}=3600$
$\Rightarrow \frac{10h^2}{3}=3600$
$\Rightarrow h^2=\frac{3600\times 3}{10}$
$\Rightarrow \sqrt{h^2}=\sqrt{1080}$
$\therefore h=32.86km$
Thus, height of tower is 32.86 km.

So, option c is correct.