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(a) 5 minutes

(b) 10 minutes

(c) $7\dfrac{1}{2}$ minutes

(d) $2\dfrac{1}{2}$ minutes

Answer

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In this question, we are supposed to find the time the boat will take to reach the shore.

So, before proceeding for this, we must know the following situation to be shown in the diagram where h is the height of the rock.

Now, from the figure it is clear that we can use the right angles triangle properties so that we can get the distance travelled by boat as:

$\tan {{30}^{\circ }}=\dfrac{h}{BC}$

So, by solving the above expression, we get:

$\begin{align}

& \dfrac{1}{\sqrt{3}}=\dfrac{h}{BC} \\

& \Rightarrow BC=h\sqrt{3} \\

\end{align}$

Now, by using the same property of the trigonometry for another triangle ABD is as:

$\tan {{60}^{\circ }}=\dfrac{h}{BD}$

So, by solving the above expression, we get:

$\begin{align}

& \sqrt{3}=\dfrac{h}{BD} \\

& \Rightarrow BD=\dfrac{h}{\sqrt{3}} \\

\end{align}$

Now, the distance travelled by boat in 15 minutes is given by the subtraction of the above two calculations as:

$h\sqrt{3}-\dfrac{h}{\sqrt{3}}$

Now, we know the formula for the speed, distance and time where v is speed, t is time and d is distance as:

$v=\dfrac{d}{t}$

So, by substituting the values of distance from above calculation and value of times as 15 minutes:

$\begin{align}

& v=\dfrac{h\sqrt{3}-\dfrac{h}{\sqrt{3}}}{15} \\

& \Rightarrow v=\dfrac{3h-h}{15\times \sqrt{3}} \\

& \Rightarrow v=\dfrac{2h}{15\sqrt{3}} \\

\end{align}$

So, to reach the shore, the boat needs to travel a distance of $\dfrac{h}{\sqrt{3}}$ in time t.

Then, it gives the value of time t as:

$\begin{align}

& t=\dfrac{\dfrac{h}{\sqrt{3}}}{\dfrac{2h}{15\sqrt{3}}} \\

& \Rightarrow t=\dfrac{h}{\sqrt{3}}\times \dfrac{15\sqrt{3}}{2h} \\

& \Rightarrow t=\dfrac{15}{2} \\

& \Rightarrow t=7\dfrac{1}{2} \\

\end{align}$

So, the time required by the boat to reach the shore is $7\dfrac{1}{2}$ minutes.

$\tan \theta =\dfrac{P}{B}$ where P is perpendicular and B is base.

$\sin \theta =\dfrac{P}{H}$ where P is perpendicular and H is hypotenuse.

$\cos \theta =\dfrac{B}{H}$ where B is base and H is hypotenuse.