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

NDA 2015 · Mathematics · Heights and Distances

The angles of elevation of the top of a tower standing on a...

The angles of elevation of the top of a tower standing on a horizontal plane from two points on a line passing through the foot of the tower at distances 49 m and 36 m are 43∘43^{{}^{\circ}} and 47∘47^{{}^{\circ}} respectively. What is the height of the tower?

Options

  1. A40 m
  2. B42 m
  3. C45 m
  4. D47 m

Correct answer

B. 42 m

Step-by-step solution

AB=h (height of  the tower) BD=36mBC=49m∠D=47∘∠C=43∘\begin{aligned} & \mathrm{AB}=\mathrm{h} \text{ (height\text{ of } the tower) } \\ & \mathrm{BD}=36 \mathrm{ m} \\ & \mathrm{BC}=49 \mathrm{ m} \\ & \angle \mathrm{D}=47^{{}^{\circ}} \\ & \angle \mathrm{C}=43^{{}^{\circ}}\end{aligned}
Now, in △ABD\triangle A B D, tan⁡47∘=h36m...(i)\tan 47^{{}^{\circ}}=\frac{\mathrm{h}}{36 \mathrm{ m}}...(i) and in △ABC\triangle \mathrm{ABC}, tan⁡43∘=h49mtan⁡(90∘−47∘)=h49\begin{aligned} & \tan 43^{{}^{\circ}}=\frac{\mathrm{h}}{49 \mathrm{ m}} \\ & \tan \left(90^{{}^{\circ}}-47^{{}^{\circ}}\right)=\frac{\mathrm{h}}{49} \end{aligned}\therefore \quad \cot 47^{{}^{\circ}}=\frac{\mathrm{h}}{49}...(ii)Multiplyingequations(i)and(ii)Multiplying equations (i) and (ii)tan⁡47∘⋅cot⁡47∘=h36×h49=1=h236×49h=6×7=42m\begin{aligned} & \tan 47^{{}^{\circ}} \cdot \cot 47^{{}^{\circ}}=\frac{\mathrm{h}}{36} \times \frac{\mathrm{h}}{49}=1=\frac{\mathrm{h}^2}{36 \times 49} \\ & \mathrm{ h}=6 \times 7=42 \mathrm{ m} \end{aligned}∴\therefore \quad Option (B) is correct
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