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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Heights and Distances

NDA 2022 · Mathematics · Heights and Distances

Paragraph: There are two points P and Q due south of a leaning...

Paragraph: There are two points PP and QQ due south of a leaning tower, which leans towards north. PP is at a distance xx and QQ is at a distance yy from the foot of the tower (x>y)(x > y). The angles of elevation of the top of the tower from PP and QQ are 15∘15^{{}^{\circ}} and 75∘75^{{}^{\circ}} respectively.
Question: If θ\theta is the inclination of the tower to the horizontal, then what is cot⁡θ\cot \theta equal to?

Options

  1. A2+3(x−y)x+y2+\frac{\sqrt{3}(x-y)}{x+y}
  2. B2−3(x−y)x+y2-\frac{\sqrt{3}(x-y)}{x+y}
  3. C2+3(x+y)x−y2+\frac{\sqrt{3}(x+y)}{x-y}
  4. D2−3(x+y)x−y2-\frac{\sqrt{3}(x+y)}{x-y}

Correct answer

D. 2−3(x+y)x−y2-\frac{\sqrt{3}(x+y)}{x-y}

Step-by-step solution

∴z=htan⁡75∘−y=x−y23⋅3−13+1−y=(x−y)2[3−13+3×3−33−3]−y=x−y2[33−3−3+39−3]−y=(x−y)[23−3]6−y=23(x−y)−3(x+y)6cot⁡θ=zh=23(x−y)−3(x+y)6×23x−y=2−3(x+y)x−y\begin{aligned} & \therefore \quad z=\frac{h}{\tan 75^{{}^{\circ}}}-y=\frac{x-y}{2 \sqrt{3}} \cdot \frac{\sqrt{3}-1}{\sqrt{3}+1}-y \\ & =\frac{(x-y)}{2}\left[\frac{\sqrt{3}-1}{3+\sqrt{3}} \times \frac{3-\sqrt{3}}{3-\sqrt{3}}\right]-y \\ & =\frac{x-y}{2}\left[\frac{3 \sqrt{3}-3-3+\sqrt{3}}{9-3}\right]-y \\ & =(x-y) \frac{[2 \sqrt{3}-3]}{6}-y=\frac{2 \sqrt{3}(x-y)-3(x+y)}{6} \\ & \cot \theta=\frac{z}{h}=\frac{2 \sqrt{3}(x-y)-3(x+y)}{6} \times \frac{2 \sqrt{3}}{x-y} \\ & =2-\frac{\sqrt{3}(x+y)}{x-y}\end{aligned}
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