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 angle of the depression of the top and foot of the other temple are \[{{30}^{\circ }}\] and \[{{60}^{\circ }}\]respectively. Find the width of the river and height of the other temple.

Answer 1

Hint: By applying \[\tan \theta \] to the respective right angled triangles we will get the height of the other temple and width of the river. We know that \[\tan \theta \] is the ratio of the opposite side to the adjacent side in the right angled triangle.

Complete step-by-step answer:
Given, the height of one temple is 50m high and from top of this temple, the angle of depression of the top and foot of other temple are \[{{30}^{\circ }}\]and \[{{60}^{\circ }}\] respectively.
We have to find the width and height of other temples.

So from above diagram, we can conclude that
\[\begin{align}
  & \angle GAB={{30}^{\circ }} \\
 & \angle GAF={{60}^{\circ }} \\
 & \angle BAE={{30}^{\circ }} \\
 & \angle FAE={{30}^{\circ }} \\
\end{align}\]
From right angled triangle ADC, we will get
\[\begin{align}
  & \tan {{30}^{\circ }}=\dfrac{x}{50} \\
 & \dfrac{1}{\sqrt{3}}=\dfrac{x}{50} \\
 & x=\dfrac{50}{\sqrt{3}} \\
 & x=\dfrac{50\sqrt{3}}{3} \\
\end{align}\]
We will get \[x=\dfrac{50\sqrt{3}}{3}\] by multiplying the both numerator and denominator with \[\sqrt{3}\]
So we get will get width of the river as \[\dfrac{50\sqrt{3}}{3}\] m
From right angled triangle AEB,
\[\begin{align}
  & \tan {{60}^{\circ }}=\dfrac{BE}{AE} \\
 & \sqrt{3}=\dfrac{\dfrac{50\sqrt{3}}{3}}{y} \\
\end{align}\]
Cancelling \[\sqrt{3}\]on both left hand side and right hand side, we will get,
\[y=\dfrac{50}{3}\]
So the height of the other temple h is
BC=AD-AE
\[h=50-\dfrac{50}{3}\]m
\[h=\dfrac{100}{3}\]m

so we get height of the other temple as \[h=\dfrac{100}{3}\]m
Note: Mostly students will confuse in specifying the angle of depression. It should be clear that the angle of depression is the angle between the horizontal line from the observer and line of sight to an object that is below the horizontal line.