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

Let The height of rock be = h

And Other Distances As Shown on the Figure

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▪ Given :-

• Time Taken by Boat to travel distance x = 10 minutes.

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▪ To Find :-

Time which Boat will take to Reach The Shore i.e.

Time Taken by Boat to cover distance y.

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▪ Solution :-

$\large \bf In \: \: \triangle\: \:ABC$

$\sf \tan30 \degree = \dfrac{h}{x + y} \\ \\ :\longmapsto \sf \frac{1}{ \sqrt{3} } = \frac{h}{x + y} \\ \\ \bf:\longmapsto x + y = h \sqrt{3} \: \: \: \: - - - - (i)$

Now ,

$\large \bf In \: \: \triangle\: \:ABD$

$\sf \tan60 \degree = \frac{h}{y} \\ \\ \sf :\longmapsto \sqrt{3} = \frac{h}{y} \\ \\ \bf:\longmapsto h = y \sqrt{3} \: \: \: \: - - - - (ii)$

Putting (ii) in (i) We Get,

$\mathtt {x + y =3y } \\ \\ :\longmapsto \mathtt{x = 2y} \\ \\ \Large\purple{ :\longmapsto  \underline {\boxed{{\bf y = \frac{x}{2} } }}}$

According To Question :

Time taken by boat to travel distance x = 10 mins

Hence ,

Time Taken by Boat to travel distance x/2 = 5 mins

So ,

The Boat will take 5 minutes to travel distance y

i.e.

The boat will take 5 minutes to reach the Shore .

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

GIVEN :

• ∠ACB= 30°, ∠ABD=60°

TO FIND :

• how soon will the boat reach the shore

CONCEPT :

• Speed is constant hence, distance is directly proportional to Time

SOLUTION :

we know that In ∆ ABC

: $\implies$Tan 30°= AB/BC

: $\implies$1/√3 = AB/BC

: $\implies$AB= BC/√3 eqn (i)

In ∆ ABD

: $\implies$Tan 60° = AB/BD

: $\implies$√3 = AB/BD

: $\implies$AB = √3 BD eqn (ii)

From the eqn (i) and (ii)

: $\implies$ BC/√3 = √3 BD

: $\implies$BC = 3 BD

We know that,

: $\implies$BC = BD+CD

: $\implies$3 BD= BD + 10

: $\implies$2 BD = 10

: $\implies$BD = 10/2

: $\implies$BD = 5

Therefore, 5 min take will the boat reach the shore

ADDITIONAL INFORMATION :

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