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 :
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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.
Step 1: Understanding the Concept:
In problems involving spheres subtending angles at an eye, the angle subtended (\(2\theta\)) is the angle between the two tangent lines from the eye to the sphere.
The observer's eye, the center of the sphere, and the point of tangency form a right-angled triangle.
The total height of the topmost point is the height of the center of the sphere plus the radius of the sphere. Step 2: Key Formula or Approach:
1. If distance from eye to center is \(d\) and radius is \(r\), then \(\sin \theta = \frac{r}{d}\), where \(2\theta\) is the subtended angle.
2. Height of center \(h = d \sin \phi\), where \(\phi\) is the angle of elevation.
3. Total Height \(H = h + r\). Step 3: Detailed Explanation:
Given radius \(r = 16 \text{ m}\).
Subtended angle \(2\theta = 60^{\circ} \Rightarrow \theta = 30^{\circ}\).
From the geometry: \(\sin 30^{\circ} = \frac{r}{d} \Rightarrow \frac{1}{2} = \frac{16}{d} \Rightarrow d = 32 \text{ m}\).
Angle of elevation of the center \(\phi = 75^{\circ}\).
Height of the center of the balloon (\(h\)):
\[ h = d \sin 75^{\circ} = 32 \sin(45^{\circ} + 30^{\circ}) \]
Using the sine sum formula \(\sin(A+B) = \sin A \cos B + \cos A \sin B\):
\[ \sin 75^{\circ} = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]
Multiply numerator and denominator by \(\sqrt{2}\):
\[ \sin 75^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4} \]
Substituting back into the height formula:
\[ h = 32 \cdot \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) = 8(\sqrt{6} + \sqrt{2}) \text{ m} \]
Height of the topmost point (\(H\)):
\[ H = h + r = 8(\sqrt{6} + \sqrt{2}) + 16 \]
Factor out 8:
\[ H = 8(\sqrt{6} + \sqrt{2} + 2) \] Step 4: Final Answer:
The height of the top most point is \(8(\sqrt{6} + \sqrt{2} + 2)\) meters.
