Multiple choice

A round balloon of radius $r$ subtends an angle $\alpha$ at the eye of an observer, while the angle of elevation of its centre is $\beta .$ Then height of the centre of the balloon is

  1. $r \sin a \ cosec \left( \dfrac { \beta } { 2 } \right)$
  2. $r \sin \beta \ cosec \left( \dfrac { \alpha} { 2 } \right)$
  3. $r \sin a \sec \left( \dfrac { \beta } { 2 } \right)$
  4. $r \sin \beta \sec \left( \dfrac { \alpha } { 2 } \right)$
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B Correct answer
Explanation

Let O be the center of the balloon and E be the eye of the observer. The angle subtended by the balloon is alpha, so the angle between the line of sight OE and the tangent is alpha/2, giving sin(alpha/2) = r/OE, or OE = r*cosec(alpha/2). Since the angle of elevation of the center is beta, the height of the center is h = OE*sin(beta) = r*sin(beta)*cosec(alpha/2).

AI explanation

Let the distance from the observer's eye to the center of the balloon be d. Viewing the balloon's radius r as the opposite side to the half-angle alpha/2, we use the sine ratio to write sin(alpha/2) = r/d, which gives d = r cosec(alpha/2). The angle of elevation to the center is beta, so the vertical height is found using h = d sin(beta). Substituting the value of d gives the final height as r sin(beta) cosec(alpha/2).

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