Two ships A and B are sailing straight away from a fixed-point O along routes such that $∠ AOB$ is always 120⁰. At a certain instance, OA=8 km, OB=6km 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.)

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$80 / \sqrt{37}$

$260 / \sqrt{37}$

$260 / 37$

$80 / 37$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{d x}{d t} = 20 kmph \\ \frac{d y}{d t} = 30 kmph \end{array}$

$\begin{array}{l} z^{2} = x^{2} + y^{2} - 2 x y \cos 120^{\circ} \\ = x^{2} + y^{2} + x y z^{2} = 8^{2} + 6^{2} + 8 (6) \\ \Rightarrow z = 2 \sqrt{37} z^{2} = x^{2} + y^{2} + x y \end{array}$
$\Rightarrow 2 z \frac{d z}{d t} = 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} + x \frac{d y}{d t} + y \frac{d x}{d t}$
$\Rightarrow 2 (2 \sqrt{37}) \frac{d z}{d t} = 16 (20) + 12 (30) + 8 (30) + 6 (20) \Rightarrow \frac{d z}{d t} = \frac{1040}{4 \sqrt{37}} = \frac{260}{\sqrt{37}} kmph$