Multiple choice

A stationary balloon is observed from $3$ points $A,\ B$ and $C$ on the plane ground and is found that its angle of elevation from each point is $\alpha$. If $\angle ABC=\beta$ and $AC=b$, the height of the balloon is

  1. ${\dfrac{b}{2}\tan\alpha}\cdot\text{cosec }\beta$
  2. $\displaystyle \frac{b}{2}\tan\alpha\cdot\sin\beta$
  3. $\displaystyle \frac{b}{2}\cot\alpha\cdot\text{cosec }\beta$
  4. $\displaystyle \frac{b}{2}\cot\alpha\cdot\sin\beta$
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A Correct answer
Explanation

Let h be the height of the balloon. The distance from the projection of the balloon on the ground to each point A, B, C is r = h * cot(alpha). Points A, B, C lie on a circle of radius r. In triangle ABC, by the law of sines, b / sin(beta) = 2r. Substituting r, we get b / sin(beta) = 2 * h * cot(alpha). Solving for h gives h = b * tan(alpha) / (2 * sin(beta)). Note that 1/sin(beta) is cosec(beta).

AI explanation

Let the balloon be at point P directly above ground point O, and let its height be h. Since the angle of elevation alpha is the same from points A, B, and C on the ground, the distance from O to A, B, and C must be identical, making O the circumcenter of triangle ABC. In the right triangle POA, tan(alpha) equals h divided by the circumradius R, so h equals R tan(alpha). According to the extended law of sines for triangle ABC, R equals b divided by (2 sin beta), which is (b/2) cosec beta. Substituting R into the height equation gives h = (b/2) tan(alpha) cosec beta.

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