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 Two ships A and B are sailing straight away from a fixed point O along routes such that \angleAOB is always 1200. At a certain instance, OA = 8 km, OB = 6 km and the ship A is sailing at the rate of 20 km/hr while the ship B sailing at the rate of 30 km/hr. Then the distance between A and B is changing at the rate (in km/hr) :

  • Option 1)

    \frac{260}{\sqrt{37}}

  • Option 2)

    \frac{260}{37}

  • Option 3)

    \frac{80}{\sqrt{37}}

  • Option 4)

    \frac{80}{37}

 

Answers (1)

As we learnt in 

 

Height and Distances -

The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

-

 In \bigtriangleup OAB, \: AB^{2}=OA^{2}+OB^{2}-2OA.OB.\cos 120^{\circ}

                               =OA^{2}+OB^{2}-2OA.OB\left ( -\frac{1}{2} \right )

                               =OA^{2}+OB^{2}+OA\cdot OB

Now, \frac{d(AB)^{2}}{dt}= \frac{d(OA^{2}+OB^{2}+OA.OB)}{dt}        [Differenting both side by t]

\Rightarrow 2AB\cdot \frac{d(AB)}{dt}=2OA\cdot \frac{d(OA)}{dt}+2OB\cdot \frac{d(OB)}{dt}+OA\cdot \frac{dB}{dt}+OB\cdot \frac{dA}{dt} \: \: \: \cdot \cdot \cdot \cdot (1)

Now, at the given instant, OA=8, OB=6       \frac{dOA}{dt}=20      .......................... (2)

                                                               \frac{dOB}{dt}=30

Now, AB= \sqrt{36+64+48}= \sqrt{148}=2\sqrt{37}               .......................... (2)

Using (1) and (2),

\frac{d(AB)}{dt}=\frac{160+180+120+60}{2\sqrt{37}}=\frac{260}{\sqrt{37}}

                                     


Option 1)

\frac{260}{\sqrt{37}}

This option is correct

Option 2)

\frac{260}{37}

This option is incorrect

Option 3)

\frac{80}{\sqrt{37}}

This option is incorrect

Option 4)

\frac{80}{37}

This option is incorrect

Posted by

Sabhrant Ambastha

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