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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: What is the length of the tower?

Options

  1. Ax−y231+{2+3(x+y)x−y}2\frac{x-y}{2 \sqrt{3}} \sqrt{1+\left\{2+\frac{\sqrt{3}(x+y)}{x-y}\right\}^2}
  2. Bx−y231+{2−3(x+y)x−y}2\frac{x-y}{2 \sqrt{3}} \sqrt{1+\left\{2-\frac{\sqrt{3}(x+y)}{x-y}\right\}^2}
  3. Cx−y431+{2+3(x+y)x−y}2\frac{x-y}{4 \sqrt{3}} \sqrt{1+\left\{2+\frac{\sqrt{3}(x+y)}{x-y}\right\}^2}
  4. Dx−y431+{2−3(x+y)x−y}2\frac{x-y}{4 \sqrt{3}} \sqrt{1+\left\{2-\frac{\sqrt{3}(x+y)}{x-y}\right\}^2}

Correct answer

B. x−y231+{2−3(x+y)x−y}2\frac{x-y}{2 \sqrt{3}} \sqrt{1+\left\{2-\frac{\sqrt{3}(x+y)}{x-y}\right\}^2}

Step-by-step solution

 Length of  the tower, AB=h2+z2=(x−y)212+(23(x−y)−3(x+y)6)2=(x−y)231+(2−3(x+y)x−y)2\begin{aligned} & \text{ Length\text{ of } the tower, } A B=\sqrt{h^2+z^2} \\ & =\sqrt{\frac{(x-y)^2}{12}+\left(\frac{2 \sqrt{3}(x-y)-3(x+y)}{6}\right)^2} \\ & =\frac{(x-y)}{2 \sqrt{3}} \sqrt{1+\left(2-\frac{\sqrt{3}(x+y)}{x-y}\right)^2}\end{aligned}
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