Question

A man on a cliff observes a boat at an angle of depression of 30 which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later the angle of depression of the boat is found to be 60.
(i) Time taken by boat to reach the shore.
(ii) Which concept is being used in this question?
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Answer 1

$\Large{\underbrace{\underline{\bf{TO\: FIND\::}}}}$

1. Tíme taken by boàt to reach the shorè.
2. Which concépt is beíng usèd?

$\Large{\underbrace{\underline{\bf{GIVEN\::}}}}$

• A man on a clíff observes a boàt at an angle of depressíon of 30°, which is approaching the shorè to the point immediately benéath the observer with a uniform speed.

• Síx minutes later the angle of depressíon of the boàt is found to be 60°.

$\Large{\underbrace{\underline{\bf{SOLUTION\::}}}}$

☛ First assume the two positíons of the boàt at the two ínstants are A & D.

Let,

• The speed of the boàt is v m/min.

• h is the height of the clíff.

• C is the locatíon of the man.

[Note ➝ The figure is given as in the attachment.]

✳ The distance covered from the point A to the point D is given as,

$\orange\bigstar\:\mid\:\bf\purple{Distance\:=\: Speed\times{Time}\:}\:\mid\:\green\bigstar$

➟ Distance (AD) = v × 6

➟ Distance (AD) = 6v

☛ Now, assume that the boàt takes t time to reach the shorè then the distance covered from the point D to the point B is given as,

➟ Distance (DB) = v × t

➟ Distance (DB) = vt

☛ Now apply the trigonometric ratio in the ∆DBC,

$\Large\purple\star$ $\mid\:\bf\pink{\tan\theta\:=\:\dfrac{Perpendicular}{Base}\:}\:\mid$ $\Large\blue\star$

➵ $\sf{\tan{60}\:=\:\dfrac{BC}{DB}\:}$

➵ $\sf{\sqrt{3}\:=\:\dfrac{h}{vt}\:}$

➵ $\sf{h\:=\:\sqrt{3}\:vt}---(1)$

☛ Now apply the trigonometric ratio in the ∆ABC,

➵ $\sf{\tan{30}\:=\:\dfrac{BC}{AB}\:}$

➵ $\sf{\dfrac{1}{\sqrt{3}}\:=\:\dfrac{h}{6v\:+\:vt}\:}$

➵ $\sf{h\:=\:\dfrac{v\:(6\:+\:t)}{\sqrt{3}}\:}---(2)$

☛ Compare the value of h from the equation (1) & equation (2),

$:\implies\:\sf{\sqrt{3}\:vt\:=\:\dfrac{v\:(6\:+\:t)}{\sqrt{3}}\:}$

$:\implies\:\sf{\sqrt{3}t\:=\:\dfrac{6\:+\:t}{\sqrt{3}}\:}$

$:\implies\:\sf{\sqrt{3}t\times{\sqrt{3}}\:=\:6\:+\:t\:}$

$:\implies\:\sf{3t\:=\:6\:+\:t\:}$

$:\implies\:\sf{3t\:-\:t\:=\:6\:}$

$:\implies\:\sf{2t\:=\:6\:}$

$:\implies\:\sf{t\:=\:\dfrac{6}{2}\:}$

$:\implies\:\bf{t\:=\:3\: minutes}$

✰ Therefore, the boàt will take 3 minutes to reach the shorè.

☛ As given that, take 6 minutes to reach the point D from the point A and we have find that the boàt takes 3 minutes to reach the shorè from the point D, thus the total time taken is,

⇒ Total time = (6 + 3) minutes

⇒ Total time = 9 minutes

∴ ⑴ The boàt will take 9 minutes to reach the shorè.

∴ ⑵ "Trigonometric ratio" concept is used in this problem.

Answer 2

Refer to the attachment for the diagram and labels !!

Given:-

• angle of depression initially = 30⁰
• angle of depression after 6 Min = 60⁰

To find:-

• time taken by the boat to reach the shore.
• concept used in this question.

Understanding the Question:-

• initially the boat was at position B. After 6 Minutes it was found to be at the position C.
• We have to find time taken by this boat to travel to the shore i.e the distance CD.

Assumption:-

• let the time taken by this boat to travel CD = t min.
• let it's constant speed = x m/min.

We know that:-

$\underline{\boxed{\bf Distance = Speed × time }}$

:$\implies$ distance BC = 6 × x

:$\implies$ distance BC = 6x m

:$\implies$ distance CD = x × t

:$\implies$ distance CD = xt m

Solution:-

We know that:-

$\underline{\boxed{\bf TanØ = \dfrac{Opposite\:Side}{Adjacent\:side}}}$

$\longrightarrow$ $\sf Tan60⁰ = \dfrac{h}{xt}$

$\longrightarrow$ $\sf √3 = \dfrac{h}{xt}$

$\longrightarrow$ $\sf h = √3xt$ ... eq.1

$\longrightarrow$ $\sf Tan30⁰ = \dfrac{h}{6x+xt}$

$\longrightarrow$ $\sf \dfrac{1}{√3} = \dfrac{h}{x (6+t)}$

$\longrightarrow$ $\sf \dfrac{x (6+t)}{√3} = h$

Now, putting the value of h in eq. 1, we get:-

$\dashrightarrow$ $\sf \dfrac{x (6+t)}{√3} = √3xt$

$\dashrightarrow$ $\sf x (6+t) = 3xt$

$\dashrightarrow$ $\sf 6+t = 3t$

$\dashrightarrow$ $\sf 2t = 6$

$\dashrightarrow$ $\sf t = 3 min$

hence, the required time is 3 minutes.

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Additional information:-

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

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