Animated Solution for Mathematics - Trigonometry: The angle of elevation of a cloud $C$ from a point $P$, $200\text{m}$ above a still lake is $30^{\circ }$. If the angle of depression of the image of $C$ in the lake from the point $P$ is $60^{\circ }$, then $PC$ (in $m$) is equal to :

Imagine you are standing on the edge of a perfectly still, glassy lake. You are positioned at a point $P$, exactly $200\text{m}$ above the water. You look up at a cloud $C$ with an angle of elevation of $30^{\circ }$, and look down at its reflection $C^{\prime }$ with an angle of depression of $60^{\circ }$.

This is a problem of symmetry, reflection, and coordinate geometry. Let us break this down step by step to uncover the hidden beauty of the setup.

The lake acts as a perfect plane mirror. In optics, a plane mirror creates an image that is as far behind the mirror as the object is in front of it.

If our cloud $C$ is at a height $h$ above the lake surface, its reflection $C^{\prime }$ must be at a depth $h$ below the lake surface. This is the fundamental anchor of our problem.

Let us visualize the geometry by drawing a horizontal line from $P$ to a point $D$ directly below the cloud. This line $PD$ represents our common horizontal distance, which we will call $x$.

In the upper triangle, $△PCD$, the vertical side $CD$ is the height of the cloud relative to the observer. Since the cloud is at height $h$ above the lake and the observer is at $200\text{m}$, the vertical distance is:

In the lower triangle, $△P{C}^{\prime }D$, the vertical side $C^{\prime }D$ is the distance from the observer down to the image. This is the distance from the observer to the lake ($200\text{m}$) plus the depth of the image below the lake ($h$):

With our triangles defined, we invoke the power of trigonometry. For the upper triangle, we use the tangent ratio:

Since $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$, we derive:

For the lower triangle, we use the angle of depression of $60^{\circ }$:

Since $\tan 60^{\circ }=\sqrt{3}$, we derive:

By equating the two expressions for $x$, we eliminate the variable and solve for $h$:

We have found the height of the cloud $h=400\text{m}$. However, the question asks for the distance $PC$, which is the hypotenuse of $△PCD$.

Given $CD=h-200=400-200=200\text{m}$, we use the sine ratio:

The final distance $PC$ is $400\text{m}$. By visualizing the physics and carefully executing the algebra, we have arrived at a precise and elegant solution.