Question

A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of the tower is ${{60}^{\circ }}$. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is ${{30}^{\circ }}$. Find the height of the tower and the width of the road.

Answer 1

Hint: Draw the diagrammatic representation of the given data/information. From the figure, consider two triangles from which the unknowns can be related to the known values using trigonometric values. Know that the tangent of an angle ($\theta $) in a triangle is given by, \[\tan \theta =\dfrac{Opposite\,\,\,side}{Adjacent\,\,\,side}\]. Use the following trigonometric values; $\tan {{60}^{\circ }}=\sqrt{3}$,$\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$

Complete step by step answer:
From the question, we can draw the following representative figure.

From the figure, we have, height of tower AB=$h$ and the width of the road BC=$x$.
We know that \[\tan \theta =\dfrac{Opposite\,\,\,side}{Adjacent\,\,\,side}\]
Now in $\Delta ABC$ from the figure, $\tan C=\dfrac{AB}{BC}$
Substituting the values of AB, BC and angle C,
$\Rightarrow \tan {{60}^{\circ }}=\dfrac{h}{x}$
We know that $\tan {{60}^{\circ }}=\sqrt{3}$.
$\Rightarrow \sqrt{3}=\dfrac{h}{x}$
On cross-multiplying, we get,
$\Rightarrow x\sqrt{3}=h...............(1)$
Also in $\Delta ABD$from the figure, $\tan D=\dfrac{AB}{BD}$
Substituting the values of AB, BD and angle D,
$\Rightarrow \tan {{30}^{\circ }}=\dfrac{h}{x+10}$
We know that $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$.
$\Rightarrow\therefore \dfrac{1}{\sqrt{3}}=\dfrac{h}{x+10}...............(2)$
Substituting the value of $h$ from equation (1) in equation (2), we have,
$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}x}{x+10}$
On cross-multiplying,
$\Rightarrow x+10=3x$
$\Rightarrow 2x=10$
$\therefore x=5\,m$
Hence, the width of the road (BC=$x$) is 5 m.
Now we also need to find the height of the tower,$h$.
Substituting $x=5\,m$ in equation (1), we obtain,
$\Rightarrow h=\sqrt{3}x$
$\therefore h=5\sqrt{3}\,m$
Hence, the height of the TV tower (AB=$h$) is $5\sqrt{3}\,m$.

$\therefore $ The height of the tower and the width of the road are $5\sqrt{3}\,m$ and 5 m respectively.

Note: The same question could have been asked keeping the angles of elevation as unknowns and to find those angles (Height would be given in this case). The method would remain the same. You have to keep in mind the difference between angle of elevation and angle of depression. Do not confuse it with each other.