Multiple choice

The angle of elevation of a stationary cloud from a point $2500$ m above a lake is $15^0$ and from the same point the angle of depression of its reflection in the lake is $45^o$. The height (in meters) of the cloud above the lake, given that $\cot\,15^o=2+\sqrt{3}$, is

  1. $2500$
  2. $2500\sqrt{2}$
  3. $2500\sqrt{3}$
  4. $5000$
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C Correct answer
Explanation

Let h be the height of the cloud. Elevation from 2500m is 15 deg. tan(15) = (h-2500)/d. Depression of reflection is 45 deg. The reflection is at -h. tan(45) = (h+2500)/d. Since tan(45)=1, d = h+2500. Substitute into first: (h-2500)/(h+2500) = tan(15) = 2-sqrt(3). Solve for h: h-2500 = (2-sqrt(3))(h+2500). h-2500 = 2h + 5000 - sqrt(3)h - 2500sqrt(3). h(sqrt(3)-1) = 7500 - 2500sqrt(3) = 2500(3-sqrt(3)) = 2500sqrt(3)(sqrt(3)-1). h = 2500sqrt(3).

AI explanation

Let the height of the cloud above the lake be h meters, making its height above the observation point (h - 2500) meters, and the depth of its reflection below the observation point (h + 2500) meters. Using the angle of elevation and depression, the horizontal distance to the cloud equals (h - 2500) cot 15 and also (h + 2500) cot 45, which is (h + 2500). Substituting the value cot 15 = 2 + sqrt(3) gives the equation (h + 2500) = (h - 2500)(2 + sqrt(3)), which simplifies to h sqrt(3) = 2500(3 + sqrt(3)). Factoring out sqrt(3) on the right side yields h sqrt(3) = 2500 sqrt(3)(sqrt(3) + 1), so the height h is 2500 sqrt(3).

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