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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Heights and Distances

NDA 2015 · Mathematics · Heights and Distances

Two poles are 10 m and 20 m high. The line joining their tops makes...

Two poles are 10 m and 20 m high. The line joining their tops makes an angle of 15∘15^{{}^{\circ}} with the horizontal. The distance between the poles is approximately equal to

Options

  1. A36.3 m
  2. B37.3 in
  3. C38.3 m
  4. D39.3 in

Correct answer

B. 37.3 in

Step-by-step solution

tan⁡15∘=DE/AEAE=10cot⁡15∘....(1)cot⁡(15∘)=cot⁡(45−30∘)=cot⁡45∘cot⁡30∘+1cot⁡30∘−cot⁡45∘\begin{aligned} & \tan 15^{{}^{\circ}}=\mathrm{DE} / \mathrm{AE} \\ & \mathrm{AE}=10 \cot 15^{{}^{\circ}} ....(1)\\ & \cot \left(15^{{}^{\circ}}\right)=\cot \left(45-30^{{}^{\circ}}\right) \\ & =\frac{\cot 45^{{}^{\circ}} \cot 30^{{}^{\circ}}+1}{\cot 30^{{}^{\circ}}-\cot 45^{{}^{\circ}}} \end{aligned}cot⁡15∘=1⋅3+13−1=2+3 Putting cot⁡15∘ in  eq (1) AE=10cot⁡15∘=10(2+3)=10(3.73)=37.3m\begin{aligned} & \cot 15^{{}^{\circ}}=\frac{1 \cdot \sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3} \\ & \text{ Putting } \cot 15^{{}^{\circ}} \text{\text{ in } eq (1) } \\ & \mathrm{AE}=10 \cot 15^{{}^{\circ}} \\ & =10(2+\sqrt{3}) \\ & =10(3.73)=37.3 \mathrm{ m} \end{aligned}$
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