A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point

Q: Why did we choose $\tan$ instead of $\sin$ or $\cos$?

Great question! We choose $\tan$ because it links the two sides we care about: the height of the tower (opposite) and the distance along the water (adjacent). $\sin$ and $\cos$ involve the hypotenuse (the line of sight), which we don't know and don't need. Always pick the ratio that connects your knowns to your unknowns!

Q: The problem doesn't give the height of the tower. How can we solve it without $h$?

That is the beauty of this problem! In many physics and math problems, you'll find that certain variables are 'intermediate'—they exist to help you build the bridge, but they cancel out in the end. Trust the algebra; if you have enough constraints, the unknown $h$ will vanish, leaving you with the relationship between distance and time.

Q: Why do we rationalize the denominator? Is it just for looks?

It's more than just aesthetics! Rationalizing the denominator is a standard mathematical practice to make the expression easier to interpret and compare. In competitive exams like JEE, the final answer is often presented in a rationalized form, so it's a habit you should build to ensure your answer matches the options provided.

Question

A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is $30^{\circ}$ (Ignore man's height). After sailing for 20 seconds towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is $45^{\circ}$. Then the time taken (in seconds) by the boat from B to reach the base of the tower is

Options:

(A) $10(\sqrt{3}-1)$

(B) $10\sqrt{3}$

(C) 10

(D) $10(\sqrt{3}+1)$

Sigma Percentile · Accepted Answer

### The Geometry of the Sea: A Journey into Trigonometry

Welcome, future engineer. Today, we are not just solving a problem; we are stepping into the shoes of an observer on a tower, watching the rhythmic dance of a boat on the water.

This problem is a classic in the JEE Advanced repertoire, not because it is computationally heavy, but because it tests your ability to translate a physical narrative into a clean, elegant mathematical model. Let us begin.

### Phase 1: Visualizing the Tower and the Boat

Imagine you are standing at the top of a tower of height $h$. You look down at the sea and see a boat at point $A$. The angle of depression is $30^{\circ}$.

Remember your geometry: the angle of depression from the top is equal to the angle of elevation from the boat, thanks to the property of alternate interior angles. So, the boat is looking up at you at $30^{\circ}$.

As the boat sails towards the tower for $20$ seconds, it reaches point $B$, where the angle of depression increases to $45^{\circ}$. As the boat gets closer to the base of the tower, you have to look down more steeply to see it. The angle of elevation from the boat to the top of the tower is now $45^{\circ}$.

### Phase 2: The Mathematical Bridge

We have two right-angled triangles here. Let the base of the tower be $Q$. The first triangle is $	riangle PBQ$, where $P$ is the top of the tower.

The angle at $B$ is $45^{\circ}$. The tangent of this angle is the ratio of the opposite side (the height of the tower, $h$) to the adjacent side (the distance $BQ$, which we will call $y$).

$$ 	an 45^{\circ} = \frac{h}{y} $$

Since $	an 45^{\circ} = 1$, we immediately find that $h = y$. This is a powerful realization; the height of the tower is exactly equal to the distance the boat has left to travel.

Now, let us look at the larger triangle, $	riangle PAQ$. The angle at $A$ is $30^{\circ}$. The tangent of this angle is the ratio of the height $h$ to the total distance $AQ$, where $AQ = x + y$.

$$ 	an 30^{\circ} = \frac{h}{x+y} $$

We know that $	an 30^{\circ} = \frac{1}{\sqrt{3}}$. Therefore:

$$ \frac{1}{\sqrt{3}} = \frac{h}{x+y} $$

### Phase 3: The Algebra of Cancellation

This is where the magic happens. We have two equations, but we do not need to know the value of $h$. We can substitute $h = y$ into our second equation:

$$ \frac{1}{\sqrt{3}} = \frac{y}{x+y} $$

Now, let us solve for $x$ in terms of $y$. Cross-multiplying gives us $x + y = y\sqrt{3}$.

Rearranging this, we get $x = y\sqrt{3} - y$, or $x = y(\sqrt{3} - 1)$. This equation tells us exactly how the distance $x$ relates to the distance $y$.

### Phase 4: The Final Countdown

We are given that the boat covers distance $x$ in $20$ seconds. Since the speed $v$ is uniform, $v = \frac{x}{20}$. We want to find the time $t$ it takes to cover the remaining distance $y$, where $t = \frac{y}{v}$.

Substituting $v = \frac{x}{20}$, we get:

$$ t = \frac{y}{\frac{x}{20}} = \frac{20y}{x} $$

Now, substitute our expression for $x$:

$$ t = \frac{20y}{y(\sqrt{3} - 1)} = \frac{20}{\sqrt{3} - 1} $$

To finalize our answer, we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{3} + 1)$:

$$ t = \frac{20(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{20(\sqrt{3} + 1)}{3 - 1} = \frac{20(\sqrt{3} + 1)}{2} $$

The final result is:

**$t = 10(\sqrt{3} + 1)$ seconds**

And there you have it! Notice how the height of the tower, which we did not even know, simply vanished from the equation. This is the elegance of physics and mathematics—sometimes, the variables that seem most important are just scaffolding for the final, beautiful result.

Select Answer:

Visualized Solution

Visualizing the Tower and Boat

Boat's Movement to Point $B$

Defining Distances

Analyzing $△ P B Q$

Height equals Remaining Distance

Analyzing $△ P A Q$

Substituting Values in $△ P A Q$

Finding $x$ in terms of $y$

Relating Distance, Speed, and Time

Time to Reach the Tower

Substituting $x$ to Find Time

Rationalizing the Denominator

Final Answer

The Sigma Insight: Heights and Distances

The Geometry of the Sea

Phase 1

Phase 2

Phase 3

Phase 4

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