Multiple choice

The angle of elevation of the top of a hill from each of the vertices$A,\ B,\ C$ of a horizontal triangle is $\alpha$. The height of the hill is where $a=BC,\ b=CA,\ c=AB$

  1. $\displaystyle \frac{1}{2} atan \alpha cosec A$
  2. $c\tan\alpha cosec C$
  3. $b\tan\alpha cosecB$
  4. $a \tan\alpha cosecA$
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A Correct answer
Explanation

The hill's projection on the horizontal plane is equidistant from A, B, and C, so its distance from each vertex is the triangle's circumradius R. Since R = a/(2 sin A), the height is R tan alpha = (1/2)a tan alpha cosec A. This matches option A.

AI explanation

Let the hill have height h and its base be point P, which is equidistant from the vertices of the triangle since the angle of elevation alpha is the same from A, B, and C. This makes P the circumcenter of triangle ABC, so the distance from P to any vertex is the circumradius R. In the right triangle formed at vertex A, tan(alpha) = h / R, which means h = R tan(alpha). By the extended law of sines, the circumradius R equals a divided by (2 sin A), or (1/2) a cosec A. Substituting this R into the height equation gives h = (1/2) a tan(alpha) cosec A.

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