At what height is the top of the tower above the ground level?

Consider the following for the next three (03) items that follow : Two points P and Q lie to the south of a leaning tower, which leans towards north. The distance of P from the foot of the tower is x and the distance of Q is y (x \gt y). The angles of elevation of the top of the tower from P and Q are 15^{\circ} and 75^{\circ} respectively.

  1. A. \frac{x-y}{2\sqrt{3}} ✓
  2. B. \frac{x-y}{4\sqrt{3}}
  3. C. \frac{x-y}{4}
  4. D. \frac{x-y}{2}

Correct Answer: A. \frac{x-y}{2\sqrt{3}}

Explanation

Let h be the height of the top, and d be the horizontal distance of its projection from the foot (towards North). Distances from P and Q to the projection are x+d and y+d. From P: \tan 15^{\circ} = \frac{h}{x+d} \implies x+d = h(2+\sqrt{3}). From Q: \tan 75^{\circ} = \frac{h}{y+d} \implies y+d = h(2-\sqrt{3}). Subtracting the equations gives x-y = h(2\sqrt{3}), which means h = \frac{x-y}{2\sqrt{3}}.

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