Spherical balloon seen under angular width — height of center A round balloon of radius r subtends an angle α at the eye. The elevation of its center from the eye is β. Find the height of the balloon’s center above the eye level.

Difficulty: Medium

Correct Answer: r Cosec α/2 Sin β

Explanation:


Introduction / Context:
The angular diameter α of a circular object fixes the distance to its center via chord geometry. The elevation β then projects this slant range to a vertical height.


Given Data / Assumptions:

  • Balloon is a sphere; apparent circle at the eye has angular diameter α.
  • Radius r known; center elevation angle = β.
  • Line of sight is straight; small-angle corrections not required.


Concept / Approach:
For a circle of radius r seen with angular diameter α, the angular radius is α/2. The geometry of a chord gives 2r = 2D sin(α/2), where D is the distance from eye to center. Thus D = r / sin(α/2) = r · csc(α/2). The center’s vertical height is then D sin β.


Step-by-Step Solution:

D = r / sin(α/2) = r · cosec(α/2)Height H = D · sin β = r · cosec(α/2) · sin β


Verification / Alternative check:
Units: r times dimensionless trig ratios → length (OK). Limiting case: as α increases (closer balloon), cosec(α/2) decreases, so D and H decrease as expected.


Why Other Options Are Wrong:
Using sin α instead of sin(α/2) ignores that α is a full angular diameter, not the angular radius. Cosec α yields the wrong scale.


Common Pitfalls:
Halving α incorrectly or treating α as an elevation angle rather than an angular size.


Final Answer:
r Cosec α/2 Sin β

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