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

Options

  1. Ax−y23\frac{x-y}{2 \sqrt{3}}
  2. Bx−y43\frac{x-y}{4 \sqrt{3}}
  3. Cx−y4\frac{x-y}{4}
  4. Dx−y2\frac{x-y}{2}

Correct answer

A. x−y23\frac{x-y}{2 \sqrt{3}}

Step-by-step solution


In △ACQtan⁡75∘=hz+y⇒z=htan⁡75∘−y\triangle A C Q \tan 75^{{}^{\circ}}=\frac{h}{z+y}\Rightarrow z=\frac{h}{\tan 75^{{}^{\circ}}}-y ...(i) In △ACPtan⁡15∘=hz+x⇒z=htan⁡15∘−x\triangle A C P \tan 15^{{}^{\circ}}=\frac{h}{z+x}\Rightarrow z=\frac{h}{\tan 15^{{}^{\circ}}}-x ...(ii) From (i) and (ii) \begin{aligned} & \frac{h}{\tan 15^{{}^{\circ}}}-\frac{h}{\tan 75^{{}^{\circ}}}=x-y \\ & \Rightarrow h\left(\frac{\sqrt{3}+1}{\sqrt{3-1}}-\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)=x-y \\ & {\left[\because\left[\begin{\text{ array }}{l} \because \tan 75^{{}^{\circ}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ \tan 75^{{}^{\circ}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \end{\text{ array }}\right]\right.} \\ & \Rightarrow h\left(\frac{3+1+2 \sqrt{3}-3-1+2 \sqrt{3}}{3-1}\right)=x-y \\ & \Rightarrow h(2 \sqrt{3})=x-y \Rightarrow h=\frac{x-y}{2 \sqrt{3}} \end{aligned}
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