A bird flies in a circle on a horizontal plane. An observer stands at a point on the ground. Suppose

Q: Wait, why do we assume the observer is outside the circle's projection? Could they be inside?

Great question! The problem explicitly states the observer is not inside the vertical projection. If they were inside, the 'minimum' angle of elevation would be $0^\circ$ (if the bird flew directly overhead) or undefined. The constraint simplifies our geometry to a standard external observer setup.

Q: Why is the projection of the midpoint $M$ perpendicular to the line connecting the observer and the center?

Think about the symmetry of the circle. The midpoint $M$ is equidistant from $P$ and $Q$ along the arc. In the ground plane, this symmetry forces the projection $M'$ to lie on the perpendicular bisector of the chord $PQ$. Since the line $OC_1$ is the axis of symmetry for the observer's view, the geometry dictates that $\Delta OC_1M'$ is a right-angled triangle.

Q: Is there a faster way to solve this without calculating $h$?

You'll notice that $h$ cancels out in the final expression for $\tan^2 \theta$. This is a classic JEE trait! You can assume $h=1$ or any convenient value to simplify your calculations, as long as you are consistent. It saves time and reduces algebraic errors.

Question

A bird flies in a circle on a horizontal plane. An observer stands at a point on the ground. Suppose $60^\circ$ and $30^\circ$ are the maximum and the minimum angles of elevation of the bird and that they occur when the bird is at the points $P$ and $Q$ respectively on its path. Let $	heta$ be the angle of elevation of the bird when it is a point on the arc of the circle exactly midway between $P$ and $Q$. Find the numerical value of $	an^2 	heta$. (Assume that the observer is not inside the vertical projection of the path of the bird.)

Sigma Percentile · Accepted Answer

### The Bird, the Sky, and the Geometry of Sight

Imagine you are standing on a vast, flat plain. Above you, a bird is tracing a perfect, invisible circle in the sky, maintaining a constant altitude $h$.

It is a serene scene, but for a physicist, it is a playground of three-dimensional geometry. We have an observer at point $O$ on the ground, watching this bird.

As the bird moves, its angle of elevation changes, dancing between a maximum of $60^\circ$ and a minimum of $30^\circ$. Our goal is to find the angle of elevation when the bird is exactly halfway between these two extremes.

### Phase 1: The 3D Visualization

First, we must ground our thoughts. The bird flies in a horizontal circle at height $h$. Let the center of this circle be $C_1$.

The observer $O$ is on the ground. The maximum elevation of $60^\circ$ occurs at the point $P$, which is the closest point on the circle to the observer. The minimum elevation of $30^\circ$ occurs at $Q$, the farthest point.

If we drop a perpendicular from the bird's path to the ground, we create a ground projection of the circle. The points $O$, $P'$, and $Q'$ (the projections of $P$ and $Q$) are collinear.

This is the first "Aha!" moment. The entire path of the bird, when viewed from above, is a circle, and the observer lies on the line extending from the diameter passing through $P$ and $Q$.

### Phase 2: The Mathematical Translation

Let the distance from the observer $O$ to the projection $P'$ be $x$. In the right-angled triangle $\Delta OP'P$, we have the angle of elevation $\alpha = 60^\circ$.

Thus, $	an 60^\circ = \frac{h}{x}$. Since $	an 60^\circ = \sqrt{3}$, we find:

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

Now, consider the farthest point $Q$. The distance from $O$ to $Q'$ is $x + d$, where $d$ is the diameter of the circle. In the right-angled triangle $\Delta OQ'Q$, the angle of elevation is $30^\circ$.

So, $	an 30^\circ = \frac{h}{x + d}$. Since $	an 30^\circ = \frac{1}{\sqrt{3}}$, we have:

$$x + d = h\sqrt{3}$$

Subtracting our expression for $x$ from this equation, we get:

$$d = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$$

The radius of the circle is $r = \frac{d}{2} = \frac{h}{\sqrt{3}}$.

### Phase 3: The Geometry of the Midway Point

Now, the bird moves to point $M$, the midpoint of the arc $PQ$. Its ground projection $M'$ is fascinating.

Because $M$ is the midpoint of the arc, its projection $M'$ lies on the circle such that the line $C_1M'$ is perpendicular to the line $OC_1$. We are now looking at a right-angled triangle in the horizontal plane: $\Delta OC_1M'$.

We know $C_1M' = r = \frac{h}{\sqrt{3}}$. We also know the distance from the observer to the center $C_1$ is:

$$OC_1 = x + r = \frac{h}{\sqrt{3}} + \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$$

Using the Pythagorean theorem for $\Delta OC_1M'$, the distance $OM'$ squared is:

$$OM'^2 = OC_1^2 + C_1M'^2 = \left(\frac{2h}{\sqrt{3}}ight)^2 + \left(\frac{h}{\sqrt{3}}ight)^2 = \frac{4h^2}{3} + \frac{h^2}{3} = \frac{5h^2}{3}$$

### The Final Elegance

We are almost there. We need the angle of elevation $	heta$ at point $M$. In the vertical triangle $\Delta OM'M$, the tangent of the angle of elevation is $	an 	heta = \frac{h}{OM'}$.

Therefore, $	an^2 	heta = \frac{h^2}{OM'^2}$. Substituting our value for $OM'^2$:

$$	an^2 	heta = \frac{h^2}{5h^2/3} = \frac{3}{5}$$

And there it is! The height $h$ cancels out, leaving us with a clean, numerical value. The final angle of elevation is:

$$	heta = \arctan\left(\sqrt{\frac{3}{5}}ight)$$

It is a beautiful reminder that in physics, when you set up your geometry correctly, the complexity often dissolves into elegant simplicity.

Visualized Solution

Visualizing the 3D Scenario

Analyzing the Closest Point $P$

Analyzing the Farthest Point $Q$

Finding the Circle's Dimensions

The Midway Point $M$

Calculating Distance $O M^{'}$

Final Value of $tan^{2} θ$

The Sigma Insight: Heights and Distances

The Bird, the Sky, and the Geometry of Sight

Phase 1

Phase 2

Phase 3

The Final Elegance

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