Question

On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 metres away from the foot of the tower, the angle of elevation of the top of bottom of the flagpole are 60° and 30° respectively. Find the height of the tower and the flagpole mounted on it.$\left[\mathrm{Use}\sqrt{3}=1.73\right]$

Answer 1

Let $OX$ be the horizontal plane, $AD$be the tower and $CD$ be the vertical flagpole.
We have:
$AB=9$ m, ∠$DBA={30}^o$ and ∠$CBA={60}^o$
Let:
$AD=h$ m and $CD\hspace{0.17em}=x$ m



In the right ∆$ABD$, we have:
$\frac{AD}{AB}=\tan {30}^o=\frac{1}{\sqrt{3}}$
⇒ $\frac{h}{9}=\frac{1}{\sqrt{3}}$
⇒ $h=\frac{9}{\sqrt{3}}=5.19$ m
Now, in the right ∆$ABC$, we have:
$\frac{AC}{BA}=\tan {60}^o=\sqrt{3}$

⇒ $\frac{h+x}{9}=\sqrt{3}$
⇒ $h+x=9\sqrt{3}$

By putting $h=\frac{9}{\sqrt{3}}$ in the above equation, we get:
$\frac{9}{\sqrt{3}}+x=9\sqrt{3}$
⇒ $x=9\sqrt{3}-\frac{9}{\sqrt{3}}$
⇒ $x=\frac{27-9}{\sqrt{3}}=\frac{18}{\sqrt{3}}=\frac{18}{1.73}=10.4$

Thus, we have:
Height of the flagpole = 10.4 m
Height of the tower = 5.19 m