What is the length of the tower?
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.
- A. \frac{x-y}{2\sqrt{3}}\sqrt{1+\{2+\frac{\sqrt{3}(x+y)}{x-y}\}^{2}}
- B. \frac{x-y}{2\sqrt{3}}\sqrt{1+\{2-\frac{\sqrt{3}(x+y)}{x-y}\}^{2}} ✓
- C. \frac{x-y}{4\sqrt{3}}\sqrt{1+\{2+\frac{\sqrt{3}(x+y)}{x-y}\}^{2}}
- D. \frac{x-y}{4\sqrt{3}}\sqrt{1+\{2-\frac{\sqrt{3}(x+y)}{x-y}\}^{2}}
Correct Answer: B. \frac{x-y}{2\sqrt{3}}\sqrt{1+\{2-\frac{\sqrt{3}(x+y)}{x-y}\}^{2}}
Explanation
The length of the tower is L = \sqrt{h^2+d^2} = h\sqrt{1+(\frac{d}{h})^2} = h\sqrt{1+\cot^2\theta}. Substituting h = \frac{x-y}{2\sqrt{3}} and \cot\theta = 2 - \frac{\sqrt{3}(x+y)}{x-y}, we obtain L = \frac{x-y}{2\sqrt{3}}\sqrt{1+\left(2-\frac{\sqrt{3}(x+y)}{x-y}\right)^2}.