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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Heights and Distances

NDA 2016 · Mathematics · Heights and Distances

A moving boat is observed from the top of a cliff of 150 m height....

A moving boat is observed from the top of a cliff of 150 m height. The angle of depression of the boat changes from 60∘60^{{}^{\circ}} to 45∘45^{{}^{\circ}} in 2 minutes. What is the speed of the boat in metres per hour?

Options

  1. A45003\frac{4500}{\sqrt{3}}
  2. B4500(3−1)3\frac{4500(\sqrt{3}-1)}{\sqrt{3}}
  3. C450034500 \sqrt{3}
  4. D4500(3+1)3\frac{4500(\sqrt{3}+1)}{\sqrt{3}}

Correct answer

B. 4500(3−1)3\frac{4500(\sqrt{3}-1)}{\sqrt{3}}

Step-by-step solution

tan⁡60∘=150x⇒x=1503Also, tan⁡45∘=150x+y⇒x+y=150⇒y=150−x=150−1503⇒y=150(3−13)= distance travelled  Speed (m/hr)=150(3−1)3×602=4500(3−1)3\begin{aligned} & \tan 60^{{}^{\circ}}=\frac{150}{x} \Rightarrow x=\frac{150}{\sqrt{3}} \\ & \text{Also, } \tan 45^{{}^{\circ}}=\frac{150}{x+y} \\ & \Rightarrow \quad x+y=150 \\ & \Rightarrow \quad y=150-x=150-\frac{150}{\sqrt{3}} \\ & \Rightarrow \quad y=150\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)=\text{ distance travelled } \\ & \quad \text{ Speed }(\mathrm{m} / \mathrm{hr})=\frac{150(\sqrt{3}-1)}{\sqrt{3}} \times \frac{60}{2}=4500 \frac{(\sqrt{3}-1)}{\sqrt{3}}\end{aligned}$$
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