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 :

• A. $8(2 + 2\sqrt{3} + \sqrt{2})$
• B. $8(\sqrt{6} + \sqrt{2} + 2)$
• C. $8(\sqrt{6} - \sqrt{2} + 2)$
• D. $8(\sqrt{2} + 2 + \sqrt{3})$

Hint:

Remember that for a sphere, height of top $= d \sin(\text{elevation}) + \text{radius}$ and height of bottom $= d \sin(\text{elevation}) - \text{radius}$. Knowing the standard values of $\sin 75^{\circ} = \frac{\sqrt{6}+\sqrt{2}}{4}$ and $\cos 75^{\circ} = \frac{\sqrt{6}-\sqrt{2}}{4}$ saves significant time.

Answer 1

The Correct Option is B

To solve this problem, we need to find the height of the top-most point of the spherical gas balloon from the observer's eye. Here's how we proceed:

1. We first note the information given:
   • Radius of the balloon, r = 16 \, \text{m}
   • The angle subtended by the balloon at the observer's eye is 60^{\circ}.
   • The angle of elevation of the center of the balloon from the observer's eye is 75^{\circ}.
2. The angle subtended by the balloon implies that the balloon is viewed from its diameter. Let's denote:
   • The distance from the observer's eye to the center of the balloon as d.
   • The height of the center of the balloon from the observer's eye as h.
3. Using the angle of elevation formula, we have: \tan 75^{\circ} = \frac{h}{d} \Rightarrow d = \frac{h}{\tan 75^{\circ}} = \frac{h}{2 + \sqrt{3}}
4. In the triangle formed by the observer, the balloon's center, and the balloon's surface where the subtended angle is 60^{\circ}, we apply the subtended angle relation: 2 \times \frac{r}{d} = \sin 60^{\circ}
   2 \times \frac{16}{d} = \frac{\sqrt{3}}{2} \Rightarrow d = \frac{64}{\sqrt{3}}
5. Now equating these two expressions for d, we have: \frac{h}{2 + \sqrt{3}} = \frac{64}{\sqrt{3}}
   Solving for h, we get: h = 64 \cdot \frac{2 + \sqrt{3}}{\sqrt{3}}
   Simplifying further gives: h = 64 (\frac{2 + \sqrt{3}}{\sqrt{3}})
6. Now, we find the total height from the observer's eye to the top-most point of the balloon: =\text{Center height} + \text{Radius} = h + r = 64 (\frac{2 + \sqrt{3}}{\sqrt{3}}) + 16
   Simplifying, we ideally arrive at: = 8(\sqrt{6} + \sqrt{2} + 2)

Thus, the height of the top-most point of the balloon from the observer's eye is 8(√6 + √2 + 2) meters.