Question

A man on the top of a rock on a seashore observes a boat coming towards it. If it takes 20 minutes for the angle of depression to change from 30 degree to 60 degree, how soon will the boat reach the shore?​

Answer 1

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

Let assume that PQ be the rock of height 'h' meters and Let P be the foot of the rock and Q be the top of rock.

Let assume that uniform speed of the boad be v m meter per minute.

Let further assume that the angle of depression of car is 30° at A and 20 minutes later, the angle of depression is 60° at B.

We know,

Distance = Speed × Time

So, Distance AB, covered by boat in 20 minutes at the speed of v meter per minute = 20v meter.

Let assume that, time taken by boat be 't' minute to reach the shore P from point B.

So, Distance BP, covered by boat in t minutes at the speed of v meter per minute = tv meter.7

Now,

$\red{\rm :\longmapsto\:In \: \triangle \: PQB}$

$\rm :\longmapsto\:tan60 \degree \: = \: \dfrac{PQ}{PB}$

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

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

Now,

$\red{\rm :\longmapsto\:In \: \triangle \: PQA}$

$\rm :\longmapsto\:tan30 \degree \: = \: \dfrac{PQ}{PA}$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} } = \dfrac{h}{20v + tv}$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3} tv}{(20+ t)v} \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3} t}{(20+ t)}$

$\rm :\longmapsto\:3t = 20 + t$

$\rm :\longmapsto\:3t - t = 20$

$\rm :\longmapsto\:2t = 20$

$\bf\implies \:t = 10 \: minutes$

Additional Information :-

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