From the top of a light-house 60 metres high with its base at the sea-level, the angle of depression

Q: Wait, why do we use the angle of elevation from the boat? Can't we just use the angle of depression inside the triangle?

That is a fantastic question! You absolutely can use the angle of depression, but you must be careful with your geometry. If you draw a horizontal line from the top of the lighthouse, the angle of depression is the angle *above* the line of sight. The angle *inside* the triangle (at the top vertex) would be $90^\circ - 15^\circ = 75^\circ$. If you use that, you would use $\tan(75^\circ) = \frac{x}{60}$. Both methods are perfectly valid! The key is to be consistent with your geometry and not mix up the adjacent and opposite sides.

Q: Why do we use $\tan(45^\circ - 30^\circ)$? Is there a faster way to find $\tan(15^\circ)$?

In JEE Advanced, precision and reliability are key. While you could memorize the value of $\tan(15^\circ)$, deriving it using the compound angle formula is a fundamental skill. It ensures you don't make a memory error and reinforces your grasp of trigonometric identities, which appear everywhere in physics and calculus. It is a robust, fail-safe method.

Q: Do I need to rationalize the denominator in the final answer?

Always look at the options provided in the question first. In competitive exams, the answer keys are often left in a specific algebraic form. If the options are un-rationalized, leave your answer in that form to save precious time! Never do extra work that the examiner hasn't asked for.

Question

From the top of a light-house 60 metres high with its base at the sea-level, the angle of depression of a boat is $15^\circ$. The distance of the boat from the foot of the light house is

Options:

(A) $\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right) 60$ metres

(B) $\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) 60$ metres

(C) $\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2$ metres

(D) none of these

Sigma Percentile · Accepted Answer

### The Lighthouse and the Geometry of the Sea

Imagine you are standing on the shore, the salt air in your lungs, looking up at a towering lighthouse. This isn't just a physics problem; it is a moment of observation.

We have a lighthouse of height $h = 60$ m, and a boat bobbing on the waves at a distance $x$ from the base. Our goal is to bridge the gap between the observer at the top and the boat at the bottom. This is the essence of trigonometry—using the known to uncover the unknown.

### Phase 1: The Geometric Bridge

Many students stumble right at the start because of the term 'angle of depression.' It sounds like a psychological state, but in geometry, it is simply a perspective. When you are at the top of the lighthouse, looking straight out at the horizon, your line of sight is horizontal.

When you look down at the boat, you tilt your head by $15^\circ$. Here is the secret: the horizontal line at the top of the lighthouse is parallel to the sea level.

When you draw the line of sight from the top to the boat, you create a transversal. By the property of alternate interior angles, the angle of elevation from the boat looking up at the lighthouse is exactly equal to the angle of depression from the top looking down.

Suddenly, the problem transforms from a confusing 'depression' scenario into a standard right-angled triangle problem with an angle of elevation of $15^\circ$. We have successfully simplified our reality.

### Phase 2: The Trigonometric Engine

Now that we have our triangle, we define our relationship. We have the opposite side (the height of the lighthouse, $60$ m) and the adjacent side (the distance of the boat, $x$). The ratio that binds these two is the tangent function:

$$ 	an(15^\circ) = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{60}{x} $$

Our mission is to isolate $x$. Rearranging this gives us:

$$ x = \frac{60}{	an(15^\circ)} = 60 \cot(15^\circ) $$

This is where the challenge lies. $15^\circ$ is not a standard angle like $30^\circ$, $45^\circ$, or $60^\circ$. We cannot simply pull the value from a table; we must construct it. This is where the beauty of the compound angle formula shines. We can express $15^\circ$ as the difference between two standard angles: $45^\circ$ and $30^\circ$.

### Phase 3: The Calculation

We invoke the compound angle identity for tangent:

$$ 	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B} $$

By substituting $A = 45^\circ$ and $B = 30^\circ$, we get:

$$ 	an(15^\circ) = \frac{	an 45^\circ - 	an 30^\circ}{1 + 	an 45^\circ 	an 30^\circ} $$

We know that $	an 45^\circ = 1$ and $	an 30^\circ = \frac{1}{\sqrt{3}}$. Substituting these values is a moment of pure algebraic satisfaction:

$$ 	an(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)(\frac{1}{\sqrt{3}})} $$

To simplify this, we multiply the numerator and the denominator by $\sqrt{3}$:

$$ 	an(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $$

### The Final Reveal

We are almost there. Remember, our equation for the distance $x$ required $\cot(15^\circ)$, which is the reciprocal of $	an(15^\circ)$. So, we simply flip our fraction:

$$ \cot(15^\circ) = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} $$

Finally, we substitute this back into our expression for $x$:

$$ x = 60 \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} ight) $$

And there it is. The distance of the boat from the lighthouse is exactly **$60 \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} ight)$ meters**.

You didn't just solve a problem; you navigated the geometry, wielded the trigonometric identities, and arrived at the truth. This is the power of the JEE mindset—taking a complex, intimidating scenario and breaking it down into elegant, solvable steps. Keep this confidence with you for the next problem!

Animated Solution for Mathematics - Trigonometry: From the top of a light-house 60 metres high with its base at the sea-level, the angle of depression of a boat is $1 5^{\circ}$ . The distance of the boat from the foot of the light house is

Select Answer:

Visualized Solution

Visualizing the Scenario

Defining Variables

Angle of Depression

Alternate Interior Angles

Applying Trigonometry

Isolating $x$

Compound Angle Formula

Substituting Known Angles

Simplifying the Expression

Calculating $cot 1 5^{\circ}$

Final Answer

The Sigma Insight: Heights and Distances

The Lighthouse and the Geometry of the Sea

Phase 1

Phase 2

Phase 3

The Final Reveal

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