If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30º and the angle

Q: Wait, why is the reflection depth $H + 25$? Shouldn't it just be $H$?

That is a classic trap! The reflection is indeed at a depth $H$ below the lake surface. However, the observer is at point $P$, which is $25\text{ m}$ *above* the lake. So, to reach the reflection from the level of $P$, you must travel $25\text{ m}$ down to the surface and then another $H$ meters down to the image. Thus, the total vertical distance is $H + 25$.

Q: Why can't we use $\sin$ or $\cos$ here?

You *could* use them, but it would make your life much harder! $\sin$ and $\cos$ involve the hypotenuse (the line of sight distance), which we don't know and don't need. The tangent function is perfect because it links the two things we care about: the vertical height and the horizontal distance $y$, which is the same for both triangles.

Q: Is the horizontal distance $y$ really the same for both the cloud and its reflection?

Yes, absolutely! The cloud, the observer, and the reflection all lie on the same vertical plane. Since the reflection is a direct projection of the cloud into the water, the horizontal distance from the observer to the cloud's vertical line is identical to the horizontal distance to the reflection's vertical line. This is the key that unlocks the problem!

Question

If the angle of elevation of a cloud from a point $P$ which is 25 m above a lake be $30^\circ$ and the angle of depression of reflection of the cloud in the lake from $P$ be $60^\circ$, then the height of the cloud (in meters) from the surface of the lake is :

Options:

(A) $42$

(B) $50$

(C) $45$

(D) $60$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a quiet, serene lakefront. You are elevated, standing on a platform exactly $25	ext{ m}$ above the water's surface.

You look up, and there is a cloud drifting lazily in the sky. You look down, and you see its reflection shimmering in the water. This is a classic physics problem that relies on the geometry of reflection.

### The Physics of Reflection

First, we must establish our coordinate system. Let the surface of the lake be our reference line, $h = 0$. We are at point $P$, which is $25	ext{ m}$ above this line.

Let the height of the cloud from the lake surface be $H$. According to the laws of reflection in a plane mirror, the image distance is equal to the object distance.

Therefore, the reflection of the cloud forms at a depth of $H$ below the lake surface. This is the fundamental physical insight that anchors our entire calculation.

### The Geometry of Sight

Now, let's visualize the triangles. We draw a horizontal line from our observation point $P$ to the vertical line passing through the cloud. Let this horizontal distance be $y$.

This $y$ is our bridge—it connects the world above the water to the world below.

Looking up at the cloud, the vertical distance from $P$ to the cloud is $H - 25$. Looking down at the reflection, the vertical distance from $P$ to the reflection is $H + 25$.

It is crucial to see that the reflection is $H$ meters below the surface, and $P$ is $25	ext{ m}$ above the surface, making the total vertical span $H + 25$.

### The Trigonometric Bridge

We are given the angle of elevation to the cloud as $30^\circ$ and the angle of depression to the reflection as $60^\circ$. Using the tangent function, we can write two elegant equations.

For the cloud (the upper triangle):

$$	an 30^\circ = \frac{H - 25}{y}$$

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

$$y = \sqrt{3}(H - 25) \quad 	ext{--- (Equation 1)}$$

For the reflection (the lower triangle):

$$	an 60^\circ = \frac{H + 25}{y}$$

Since $	an 60^\circ = \sqrt{3}$, we have:

$$y = \frac{H + 25}{\sqrt{3}} \quad 	ext{--- (Equation 2)}$$

### The Algebraic Resolution

Now, we equate our two expressions for $y$ because the horizontal distance is the same in both scenarios:

$$\sqrt{3}(H - 25) = \frac{H + 25}{\sqrt{3}}$$

To clear the fraction, multiply both sides by $\sqrt{3}$:

$$3(H - 25) = H + 25$$

$$3H - 75 = H + 25$$

Now, bring the $H$ terms to one side and the constants to the other:

$$3H - H = 25 + 75$$

$$2H = 100$$

$$H = 50	ext{ m}$$

### Final Conclusion

And there we have it! The height of the cloud from the lake surface is **$50	ext{ m}$**.

It is a beautiful example of how a complex-looking problem in trigonometry and optics can be reduced to simple, elegant algebra. Always remember: visualize the geometry first, identify the common variables, and the math will follow naturally.

Select Answer:

Visualized Solution

Visualizing the Scene

The Cloud and its Reflection

Creating the Reference Framework

Angle of Elevation

Applying Tan in the First Triangle

Angle of Depression

Applying Tan in the Second Triangle

Equating the Two Expressions

Simplifying the Equation

Solving for $H$

Final Answer

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Physics of Reflection

The Geometry of Sight

The Trigonometric Bridge

The Algebraic Resolution

Final Conclusion

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