A tower stands at the centre of a circular park. A and B are two points on the boundary of the park

Q: Wait, why is the triangle $\Delta OAB$ equilateral? I thought it was just isosceles.

That is a fantastic observation! You are correct that $\Delta OAB$ is isosceles because $OA = OB = R$. However, in any isosceles triangle, if the vertex angle is $60^\circ$, the base angles must be $\frac{180^\circ - 60^\circ}{2} = 60^\circ$. Since all three angles are $60^\circ$, the triangle is, by definition, equilateral. This means $OA = OB = AB = a$.

Q: Why can't we use the sine rule in the vertical triangle $\Delta OAC$?

You absolutely can! The sine rule is a powerful tool. In $\Delta OAC$, we have $\frac{h}{\sin(30^\circ)} = \frac{OA}{\sin(60^\circ)}$. Since $OA = a$, you get $h = a \cdot \frac{\sin(30^\circ)}{\sin(60^\circ)} = a \cdot \frac{1/2}{\sqrt{3}/2} = \frac{a}{\sqrt{3}}$. Both methods lead to the same beautiful result.

Q: Does the position of $A$ and $B$ on the circle matter for the height of the tower?

It does not! As long as the chord $AB$ subtends $60^\circ$ at the center, the distance from the center to the chord is fixed by the geometry of the circle. The height of the tower depends only on the distance from the foot of the tower to the point of observation. Since $OA = OB = R = a$, the distance is constant regardless of where $A$ and $B$ are placed on the boundary.

Question

A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $AB (= a)$ subtends an angle of $60^\circ$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $30^\circ$. The height of the tower is

Options:

(A) $a/\sqrt{3}$

(B) $a\sqrt{3}$

(C) $2a/\sqrt{3}$

(D) $2a\sqrt{3}$

Sigma Percentile · Accepted Answer

### The Art of Geometric Decomposition

In the world of JEE Advanced physics and mathematics, the most daunting problems are often those that appear to exist in three dimensions. When you first read a problem about a tower standing in a circular park, your mind might try to construct a complex 3D model, complete with angles of elevation and chord lengths.

But here is the secret: the master key to these problems is not complex 3D calculus, but the elegant art of decomposition.

### Phase 1: The Ground Plane

Imagine you are standing directly above the park, looking down. The tower is just a point, $O$, at the center. The points $A$ and $B$ are on the boundary. This is your ground plane.

We are given that the chord $AB$ has a length of $a$. We are also told that this chord subtends an angle of $60^\circ$ at the center $O$.

This is where the magic happens. We know that $OA$ and $OB$ are both radii of the circular park, so $OA = OB = R$. This immediately tells us that $\Delta OAB$ is an isosceles triangle.

But we have a gift: the vertex angle $\angle AOB$ is $60^\circ$. As we discussed in our FAQs, this forces the base angles to also be $60^\circ$.

Suddenly, the complexity collapses. Our triangle $\Delta OAB$ is not just any triangle; it is an equilateral triangle. This means $OA = OB = AB = a$. We have just found the distance from the center of the park to the points $A$ and $B$ without doing a single calculation!

### Phase 2: The Vertical Ascent

Now that we have mastered the ground plane, let us lift our gaze to the vertical plane. We are looking for the height of the tower, $h$. Let the top of the tower be $C$.

We have a right-angled triangle $\Delta OAC$, where $O$ is the foot of the tower, $C$ is the top, and $A$ is the point on the boundary. The angle of elevation from $A$ to $C$ is given as $30^\circ$.

This is a classic right-angled triangle problem. We have the angle $\angle OAC = 30^\circ$, the opposite side $OC = h$, and the adjacent side $OA = a$. The relationship is simple and elegant:

$$	an(30^\circ) = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{OC}{OA} = \frac{h}{a}$$

### Phase 3: The Final Synthesis

We know from our trigonometric toolkit that $	an(30^\circ) = \frac{1}{\sqrt{3}}$. Substituting this into our equation, we get:

$$\frac{1}{\sqrt{3}} = \frac{h}{a}$$

Solving for $h$, we find:

$$h = \frac{a}{\sqrt{3}}$$

### The Takeaway

Look at how we arrived here. We didn't need to struggle with complex 3D coordinates or heavy vector algebra. By simply separating the problem into a 2D ground plane and a 2D vertical plane, we turned a terrifying 3D problem into a simple, solvable geometry exercise.

This is the essence of JEE Advanced preparation: it is not about memorizing formulas, but about developing the intuition to break down complex systems into their simplest, most beautiful components. Keep this mindset, and you will find that no problem is too large to conquer. The final result is **$h = \frac{a}{\sqrt{3}}$**.

Select Answer:

Visualized Solution

Visualizing the 3D Setup

The Ground Triangle $Δ O A B$

Subtended Angle $∠ A O B = 6 0^{\circ}$

Properties of $Δ O A B$

$Δ O A B$ is Equilateral

The Vertical Triangle $Δ O A C$

Angle of Elevation $∠ O A C = 3 0^{\circ}$

Applying Trigonometry in $Δ O A C$

Calculating the Height $h$

Final Answer & Key Takeaway

The Sigma Insight: Heights and Distances

The Art of Geometric Decomposition

Phase 1

Phase 2

Phase 3

The Takeaway

Similar Questions

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