question_answer

A moving boat is observed form the top of a cliff of 150 m height. The angle of depression of the boat changes form \[60{}^\circ \] to \[45{}^\circ \] in 2 minutes. What is the speed of the boat in meters per hours?

A) \[\frac{4500}{\sqrt{3}}\]

B) \[\frac{4500(\sqrt{3}-1)}{\sqrt{3}}\]

C) \[4500\sqrt{3}\]

D) \[\frac{4500(\sqrt{3}+1)}{\sqrt{3}}\]

Correct Answer: B

Solution :

[b]

\[\tan 60{}^\circ =\frac{150}{x}\Rightarrow x=\frac{150}{\sqrt{3}}\]

Also,  \[\tan 45{}^\circ =\frac{150}{x+y}\]

\[\Rightarrow x+y=150\]

\[\Rightarrow y=150-x=150-\frac{150}{\sqrt{3}}\]

\[\Rightarrow y=150\left( \frac{\sqrt{3}-1}{\sqrt{3}} \right)=\] distance travelled

Speed in (m/hr) \[=\frac{150(\sqrt{3}-1)}{\sqrt{3}}\times \frac{60}{2}\]

\[=4500\frac{(\sqrt{3}-1)}{\sqrt{3}}\]