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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)


  • Option 2)


  • Option 3)


  • Option 4)



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)


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

Using (1) and (2),



Option 1)


This option is correct

Option 2)


This option is incorrect

Option 3)


This option is incorrect

Option 4)


This option is incorrect

Posted by

Sabhrant Ambastha

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