Question

A moving boat is observed from the top of a 150 m high cliff moving away from the cliff The angle of depression of the boat changes from \[60^\circ\] to \[45^\circ\] in 2 minutes Find the speed of the boat in $\dfrac{m}{h}$.

Answer 1

Hint: In this problem, first we need to find the distance travelled by boat in 2 minutes using trigonometric identities. Next, find the speed of the boat using a time and distance formula.

Complete step-by-step solution -
The given problem can be drawn as shown below.

From the above figure, it can be concluded that,
\[\begin{gathered}
  \angle EAC = \angle ACB = 60^\circ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Alternate angle}}} \right) \\
  \angle EAD = \angle ADB = 45^\circ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Alternate angle}}} \right) \\
\end{gathered}\]
Now, In \[\Delta ABC\],
\[ \tan {60^0} = \dfrac{{AB}}{{BC}} \\
   \Rightarrow \tan {60^0} = \dfrac{{150}}{x} \\
   \Rightarrow \sqrt 3 = \dfrac{{150}}{x} \\
   \Rightarrow x = \dfrac{{150}}{{\sqrt 3 }} \\
   \Rightarrow x = \dfrac{{150}}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} \\
   \Rightarrow x = 50\sqrt 3 \,\,m \\ \]
Further, in \[\Delta ABD\],
\[\tan {45^0} = \dfrac{{AB}}{{BD}} \\
   \Rightarrow \tan {45^0} = \dfrac{{150}}{{y + x}} \\
   \Rightarrow y + x = 150 \\
   \Rightarrow y = 150 - x \\
\Rightarrow y = 150 - 50\sqrt 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [ \because\left( {x = 50\sqrt 3 } \right) ] \\
   \Rightarrow y = 50\left( {3 - \sqrt 3 } \right) \\ \]
The given time is converted into hours as shown below.
\[ 1{\text{minute}} \to \dfrac{1}{{60}}{\text{hours}} \\
   \Rightarrow 2{\text{minute}} \to \dfrac{2}{{60}}{\text{hours}} \\ \]
Now, the speed \[S\] of the boat is calculated as follows:
\[ S = \dfrac{y}{{\dfrac{2}{{60}}}}{\text{m/hr}} \\
\Rightarrow S = \dfrac{{50\left( {3 - \sqrt 3 } \right)}}{{\dfrac{1}{{30}}}}{\text{m/hr}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [ \because \left( {y = 50\left( {3 - \sqrt 3 } \right)} \right) ] \\
   \Rightarrow S = 1500\left( {3 - \sqrt 3 } \right){\text{m/hr}} \\ \]
Thus, the speed of the boat is \[1500\left( {3 - \sqrt 3 } \right){\text{m/hr}}\].

Note: Convert the time from minutes to hours in order to obtain the speed of the boat into meters per hour. The key concept of this type of problem is the tangent formula of given angles. Here we can also use the similarity property of the triangles and find the ratio of sides.