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:

Let the observer’s eye be the point $$A$$. We draw the vertical plane that contains the centre of the balloon; in this plane every relevant point lies in one simple two-dimensional diagram.

Denote by $$C$$ the centre of the spherical balloon and by $$r$$ its radius. Given data:

$$$r = 16\ \text{m},\qquad \angle(\text{elevation of }C)=75^\circ ,\qquad \text{angle subtended by the balloon}=60^\circ .$$$

Through the eye $$A$$ two lines are drawn which just touch the sphere; they meet the sphere at the upper and lower tangent points $$P$$ and $$Q$$. Because the tangents are symmetric about the line $$AC$$, the angle between $$AC$$ and each tangent is exactly one-half of the total subtended angle:

$$\angle PA C=\angle Q A C =\dfrac{60^\circ}{2}=30^\circ .$$

At the point of contact between a tangent and a sphere, the radius is perpendicular to the tangent. Hence in the right-angled triangle $$\triangle APC$$ we know

$$$\angle P = 90^\circ,\qquad \angle A = 30^\circ,\qquad CP = r = 16\ \text{m}.$$$

First we relate the distance $$AC$$ to the radius $$r$$. Stating the definition of sine in a right triangle: $$$\sin(\text{angle at }A)=\dfrac{\text{side opposite that angle}}{\text{hypotenuse}}.$$$

Here the side opposite $$\angle A$$ is $$CP$$ and the hypotenuse is $$AC$$, so

$$$\sin 30^\circ = \dfrac{CP}{AC}\; \Longrightarrow\; \dfrac12=\dfrac{16}{AC}\; \Longrightarrow\; AC = 2\times 16 = 32\ \text{m}.$$$

Now we split this line $$AC$$ into its horizontal and vertical components. Let

$$$d = (\text{horizontal distance from }A\text{ to }C),\qquad h = (\text{vertical height of }C\text{ above the observer’s eye}).$$$

The given angle of elevation is $$75^\circ$$, so by elementary trigonometry

$$\sin 75^\circ = \dfrac{h}{AC},\qquad \cos 75^\circ = \dfrac{d}{AC}.$$

Using $$AC = 32\ \text{m}$$ we obtain

$$h = 32\sin 75^\circ .$$

We next write $$\sin 75^\circ$$ in its exact surd form. Using the half-angle identity $$\cos15^\circ = \dfrac{\sqrt6+\sqrt2}{4}$$ and the co-function relation $$\sin75^\circ = \cos15^\circ$$, we have

$$\sin 75^\circ = \dfrac{\sqrt6+\sqrt2}{4}.$$

Substituting this in the expression for $$h$$:

$$$h = 32 \times \dfrac{\sqrt6+\sqrt2}{4} = 8(\sqrt6+\sqrt2)\ \text{metres}.$$$

The top-most point of the balloon, call it $$T$$, is a further radius $$r=16$$ metres above the centre. Therefore its height above the observer’s eye level is

$$$AT_{\text{(vertical)}} = h + r = 8(\sqrt6+\sqrt2) + 16 = 8(\sqrt6+\sqrt2) + 8\times 2 = 8(\sqrt6 + \sqrt2 + 2)\ \text{metres}.$$$

Hence, the correct answer is Option B.