Question

On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point 9m away from the foot of the tower the angle of elevation of the top and the bottom of the flagpole are $60{}^\circ $ and $30{}^\circ $ respectively. Find the height of the tower, and the flagpole mounted on top of it.

Answer 1

Hint: Use the fact that $\tan \alpha =\dfrac{AB}{AD}$ to find the value of AD. Use $\tan \beta =\dfrac{AC}{AD}$ to find the length of AC. Hence find the length of the flagpole using BC= AC – AB.

Complete step-by-step answer:

Given: AB is a tower. On top of the tower, a flagpole BC is placed. The angles of elevation to the bottom (B) and the top (C) of the flagpole are $\alpha =30{}^\circ $ and $\beta =60{}^\circ $. AD = 9m
To determine: The length of the flagpole (BC) and the height of the tower (AB).
In triangle ABD, we have AB is the side opposite to $\alpha $ and AD is the side adjacent to $\alpha $. We know that in a triangle
$\tan \theta =\dfrac{\text{Opposite side}}{\text{Adjacent side}}$
Hence, we have $\tan \alpha =\dfrac{AB}{AD}$
Multiplying both sides by AD, we get
$AB=AD\tan \alpha \text{ }\left( i \right)$
In triangle ACD, we have AC is the side opposite to $\beta $ and AD is the side adjacent to $\beta $
We know that in a triangle
$\tan \theta =\dfrac{\text{Opposite side}}{\text{Adjacent side}}$
Hence, we have $\tan \beta =\dfrac{AC}{AD}$
Multiplying both sides by AD, we get
$AC=AD\tan \beta \text{ }\left( ii \right)$
Subtracting equation (i) from equation (ii), we get
$AC-AB=AD\left( \tan \beta -\tan \alpha \right)$
We know that AC – AB = BC
Hence, we have
$BC=AD\left( \tan \beta -\tan \alpha \right)$
Substituting AD = 9, $\beta =60{}^\circ $ and $\alpha =30{}^\circ $, we get
$BC=9\left( \tan 60{}^\circ -\tan 30{}^\circ \right)$
We know that $\tan 60{}^\circ =\sqrt{3},\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$
Hence, we have
$BC=9\left( \sqrt{3}-\dfrac{1}{\sqrt{3}} \right)=9\left( \dfrac{3-1}{\sqrt{3}} \right)=6\sqrt{3}$
Also from equation (i), we have
$AB=AD\tan \alpha =9\times \dfrac{1}{\sqrt{3}}=3\sqrt{3}$
Hence, the height of the tower is $3\sqrt{3}m$, and the length of the flagpole is $6\sqrt{3}m$.

Note: Verification:
We have $AB=3\sqrt{3}$ and AD = 9m.
Hence, we have $\dfrac{AB}{AD}=\dfrac{3\sqrt{3}}{9}=\dfrac{1}{\sqrt{3}}=\tan 30{}^\circ $
Hence, we have $\alpha =30{}^\circ $
Also, we have AC = AB + BC $=3\sqrt{3}+6\sqrt{3}=9\sqrt{3}$
Hence, we have $\dfrac{AC}{AD}=\dfrac{9\sqrt{3}}{9}=\sqrt{3}=\tan 60{}^\circ $
Hence, we have $\beta =60{}^\circ $
Hence our answer is verified to be correct.