If \theta is the inclination of the tower to the horizontal, then what is \cot\theta equal to?
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. 2+\frac{\sqrt{3}(x-y)}{x+y}
- B. 2-\frac{\sqrt{3}(x-y)}{x+y}
- C. 2+\frac{\sqrt{3}(x+y)}{x-y}
- D. 2-\frac{\sqrt{3}(x+y)}{x-y} ✓
Correct Answer: D. 2-\frac{\sqrt{3}(x+y)}{x-y}
Explanation
The inclination of the tower \theta satisfies \cot\theta = \frac{d}{h}. Adding the equations from the previous solution: (x+d) + (y+d) = h(2+\sqrt{3}) + h(2-\sqrt{3}) = 4h. This gives 2d = 4h - (x+y), so d = 2h - \frac{x+y}{2}. Then \cot\theta = \frac{d}{h} = 2 - \frac{x+y}{2h}. Substituting 2h = \frac{x-y}{\sqrt{3}}, we get \cot\theta = 2 - \frac{\sqrt{3}(x+y)}{x-y}.