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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Heights and Distances

NDA 2014 · Mathematics · Heights and Distances

From an aeroplane above a straight road the angles of depression of...

From an aeroplane above a straight road the angles of depression of two positions at a distance 20 m apart on the road are observed to be 30∘30^{{}^{\circ}} and 45∘45^{{}^{\circ}}. The height of the aeroplane above the ground is:

Options

  1. A103m10 \sqrt{3} \mathrm{ m}
  2. B10(3−1)m10(\sqrt{3}-1) \mathrm{m}
  3. C10(3+1)m10(\sqrt{3}+1) \mathrm{m}
  4. D20 m

Correct answer

C. 10(3+1)m10(\sqrt{3}+1) \mathrm{m}

Step-by-step solution


 In △ABD,tan⁡45∘=hx⇒1=hx⇒h=x\begin{array}{ll}\text{ In } \triangle \mathrm{ABD}, & \tan 45^{{}^{\circ}}=\frac{h}{x} \\ \Rightarrow & 1=\frac{h}{x} \Rightarrow h=x\end{array}In △\triangle\mathrm{ABC}, tan⁡30∘=hx+20⇒13=hx+20=hh+20(\becausex=h)⇒3h=h+20=(3h−h)=20⇒h(3−1)=20⇒h=203−1×3+13+1=20(3+1)2⇒h=10(3+1)\begin{array}{rlrl}& & \tan 30^{{}^{\circ}} & =\frac{h}{x+20}\\ & \Rightarrow \quad \frac{1}{\sqrt{3}}& =\frac{h}{x+20}=\frac{h}{h+20} \quad(\becausex=h) \\ \Rightarrow & \quad \sqrt{3}h & =h+20=(\sqrt{3}h-h)=20 \\ \Rightarrow & h(\sqrt{3}-1) & =20 \\ \Rightarrow \quad & h & =\frac{20}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{20(\sqrt{3}+1)}{2}\\ \Rightarrow \quad & h & =10(\sqrt{3}+1) \end{array} Hence, height of the aeroplane above the ground =10(3+1)m=10(\sqrt{3}+1) \mathrm{m}.
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