A spherical gas balloon of radius 16 meter subtends an angle 60° at the eye of the observer A while

Q: Wait, why do we treat the balloon as a 2D circle? Isn't it a 3D sphere?

That is a brilliant observation! While the balloon is indeed a 3D sphere, the observer's eye, the center of the sphere, and the points of tangency all lie on a single plane. By looking at this cross-sectional plane, we reduce the complexity of the 3D geometry into a manageable 2D trigonometry problem without losing any physical accuracy.

Q: Why is the angle $\angle OAP$ equal to $30^\circ$? Is that always true?

Yes, it is! Because the two lines of sight from the observer to the sphere are tangents, they are equal in length. The line connecting the observer to the center of the sphere acts as an axis of symmetry, bisecting the total angle subtended by the sphere. Since the total angle is $60^\circ$, the half-angle must be $30^\circ$.

Q: Can I just use $\sin(75^\circ) \approx 0.966$ to get the answer?

In a JEE Advanced context, never rely on decimal approximations unless explicitly asked. The options are in radical form, which implies the examiner wants you to demonstrate mastery of trigonometric identities like $\sin(A+B)$. Using exact values ensures you reach the precise algebraic form required by the options.

Q: Why do we add the radius $r$ at the end? Is the topmost point not just $OR$?

Think of $OR$ as the height of the 'heart' of the balloon—its center. The balloon extends $r$ meters above its center and $r$ meters below it. To find the height of the very top of the balloon relative to the eye level, you must add that extra radius distance to the height of the center.

Q: Is there a risk of confusion between the angle of elevation of the center and the angle of elevation of the top point?

That is a classic trap! The problem specifically gives the angle of elevation of the *center*. If it had given the angle of elevation of the top point, the geometry would have been much more complex. Always read the problem statement carefully to identify exactly which point the angle of elevation refers to.

Question

A spherical gas balloon of radius 16 meter subtends an angle $60^\circ$ at the eye of the observer $A$ while the angle of elevation of its center from the eye of $A$ is $75^\circ$. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is:

Options:

(A) $8(2 + 2\sqrt{3} + \sqrt{2})$

(B) $8(\sqrt{6} + \sqrt{2} + 2)$

(C) $8(\sqrt{2} + 2 + \sqrt{3})$

(D) $8(\sqrt{6} - \sqrt{2} + 2)$

Sigma Percentile · Accepted Answer

### The Geometry of Sight: Unveiling the Balloon

Imagine you are standing on a vast, open field. You look up and see a massive spherical balloon floating in the sky. As a student of physics, your mind immediately begins to map the geometry of the scene, transforming a serene moment into a problem of trigonometry and spatial reasoning.

### Phase 1: The Symmetry of Tangents

When you look at a sphere from a point $A$, your lines of sight are essentially tangents to that sphere. The problem states the balloon subtends an angle of $60^{\circ}$ at your eye. If we draw a line from your eye $A$ to the center of the balloon $O$, this line acts as an axis of symmetry, bisecting the $60^{\circ}$ angle into two $30^{\circ}$ angles.

Consider the right-angled triangle formed by your eye $A$, the center $O$, and a point of tangency $P$. Because the radius $OP$ is perpendicular to the tangent $AP$, we have a right triangle $	riangle OAP$. With the half-angle $\angle OAP = 30^{\circ}$ and the radius $OP = 16$ m, we find the distance to the center using the sine ratio:

$$\sin(30^{\circ}) = \frac{OP}{OA}$$

Since $\sin(30^{\circ}) = \frac{1}{2}$, we find that $OA = 32$ m. You have successfully calculated the distance to the heart of the balloon.

### Phase 2: The Elevation of the Center

Next, we account for the angle of elevation. The center of the balloon $O$ is at an angle of $75^{\circ}$ above your horizontal eye level. Let $R$ be the point on your horizontal level directly beneath the center $O$. We now have a new right triangle $	riangle ARO$, where the hypotenuse is the distance $OA = 32$ m.

The vertical height of the center, $OR$, is given by:

$$OR = OA \sin(75^{\circ})$$

To solve this, we use the compound angle identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$ to expand $\sin(75^{\circ})$ as $\sin(45^{\circ} + 30^{\circ})$:

$$\sin(75^{\circ}) = \sin(45^{\circ})\cos(30^{\circ}) + \cos(45^{\circ})\sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$$

Substituting this back into our equation for $OR$:

$$OR = 32 \cdot \frac{\sqrt{3}+1}{2\sqrt{2}} = 16 \cdot \frac{\sqrt{3}+1}{\sqrt{2}} = 8\sqrt{2}(\sqrt{3}+1) = 8\sqrt{6} + 8\sqrt{2}$$

### Phase 3: Reaching the Summit

We have the height of the center, but the question asks for the height of the topmost point of the balloon. Since the balloon is a sphere of radius $r = 16$ m, the top point $T$ sits exactly $16$ m above the center $O$. Therefore, the total height $H$ is:

$$H = OR + r = (8\sqrt{6} + 8\sqrt{2}) + 16$$

By factoring out the $8$, we arrive at the elegant final form:

**$$H = 8(\sqrt{6} + \sqrt{2} + 2)$$**

This result is the culmination of understanding symmetry, trigonometric identities, and spatial visualization. By maintaining this clarity of thought, you ensure that no complex geometry problem will stand in your way.

Select Answer:

Visualized Solution

Visualizing the Setup

Symmetry and Half Angle

Distance to Center $O A$

Calculating $O A$

Angle of Elevation of Center

Height of Center $O R$

Calculating $sin (7 5^{\circ})$

Evaluating $O R$

Height of the Topmost Point

Final Simplification

The Sigma Insight: Heights and Distances

The Geometry of Sight

Phase 1

Phase 2

Phase 3

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