Question

A man on the top of a vertical lighthouse observes a boat coming directly towards it. If it takes 10 minutes for the angle of depression to change from 30° to 60° ,how soon it will reach the lighthouse?​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that AB be the light house of height h m and B be the top of light house.

Let assume that uniform speed of boat be v m/minute.

Let C be the initial position of the boat when angle of depression is 30° and in 10 minutes, Let D be the position of boat at which angle of depression is 60°.

Now, Distance, CD = Speed × time = 10v meter.

Let assume that boat take t minutes more to reach the lighthouse from D.

So, Distance, DA = tv meter.

Now, In right angle triangle ABD

$\rm \: tan60 \degree \: = \: \dfrac{AB}{AD}$

$\rm \: \sqrt{3} \: = \: \dfrac{h}{tv}$

$\rm\implies \:h \: = \: \sqrt{3} \: tv - - - (1)$

Now, In right angle triangle ABC

$\rm \: tan30 \degree \: = \: \dfrac{AB}{AC}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{h}{10v + tv}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3} \: tv }{(10 + t)v}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3} \: t }{10 + t}$

$\rm \: 3t = 10 + t$

$\rm \: 2t = 10$

$\rm\implies \:t \: = \: 5 \: minutes$

So, it means boat will take 5 minutes more from D to reach the lighthouse.

$\rule{190pt}{2pt}$

$\color{green}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$