Question

There are two temples, one on each bank of a river, just opposite to each other. One temple is 50m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30 and 60 degrees respectively. Find the width of the river and the height of the other temple.

Answer 1

Hint: First we will draw the diagram and then with the help of that we try to find the width and height of the river. We will first find the value of the width of the river using the formula $\tan \theta =\dfrac{height}{base}$ and then using that width we will find the value of height of the tower.

Complete step-by-step answer:
Let’s start solving this question.
First we will look at the figure,

From the above figure we can say that,
$\angle FBD=30{}^\circ$ which is the angle of depression from one top of the temple to the other top.
$\angle FBC=60{}^\circ$ which is the angle of depression from one top temple to the foot of another temple.
DC = y = height of the temple that we need to find,
AB = 50m is the height of the temple that is given in the question.
BG = h is the difference in height between the two temples.
AC = x is the width of the river that we also need to find.
In triangle ABC we will use the formula of tan,
$\tan \theta =\dfrac{height}{base}$
The $\angle FBC\text{ and }\angle BCA$ are alternate angles hence they must be equal to 60 degree.
Now we have $\theta =60{}^\circ$, height = 50m and let base = x
Therefore we get,
$\begin{aligned} & \tan 60=\dfrac{50}{x} \\\ & \sqrt{3}=\dfrac{50}{x} \\\ & x=\dfrac{50}{\sqrt{3}} \\\ \end{aligned}$
Hence, x is the width of the river which is $\dfrac{50}{\sqrt{3}}m$
Now $\angle FBD\text{ and }\angle \text{BDG}$ are alternate angles and hence they must be equal to 30 degree.
In triangle BDG we have $\theta =30{}^\circ$, height = h, and base = DG = AC = x
Now using the formula $\tan \theta =\dfrac{height}{base}$ we get,
$\begin{aligned} & \tan 30=\dfrac{h}{x} \\\ & \dfrac{1}{\sqrt{3}}=\dfrac{h}{x} \\\ \end{aligned}$
Now substituting the value of x as $\dfrac{50}{\sqrt{3}}m$ we get,
$h=\dfrac{\dfrac{50}{\sqrt{3}}}{\sqrt{3}}=\dfrac{50}{3}$m
Now the height of the tower DC or y = 50 – h
Now substituting the value of $h=\dfrac{50}{3}$ we get,
$y=50-\dfrac{50}{3}=\dfrac{100}{3}m$
Hence, the height of the tower is $\dfrac{100}{3}m$

Note: The figure that we have drawn is very important and must be drawn carefully while understanding each word given in the question. Many students make incorrect figures and due to which they get the wrong answer like in the case of the angle of depression. The angle of depression is when an observer looks at an object that is situated at a distance lower than the observer, an angle is formed below the horizontal line drawn with the level of the eye of the observer and line joining object with the observer’s eye. This must be kept in mind to avoid the mistake in making a figure.