Two ships A and B are sailing away from a fixed point O along routes such that $∠ A O B$ is always $120^{\circ}$ 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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$\frac{260}{\sqrt{37}}$

$\frac{260}{37}$

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

$\frac{80}{37}$

answer is A.

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Detailed Solution

$Let O A = x km, O B = y km, A B = R$

$R^{2} = x^{2} + y^{2} - 2 x y (- \frac{1}{2}) = x^{2} + y^{2} + x y .$

$R = a t \Rightarrow x = 6 km, and y = 8 km$

$R = \sqrt{6^{2} + 8^{2} + 6 \times 8} = 2 \sqrt{37}$

$2 R \frac{d R}{d t} = 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} + (x \frac{d y}{d t} + y \frac{d x}{d t})$

$= \frac{1}{2 R} [2 \times 8 \times 20 + 2 \times 6 \times 30 + (8 \times 30 + 6 \times 20)]$

$\Rightarrow \frac{d R}{d t} = \frac{1}{2 \times 2 \sqrt{37}} [1040] = \frac{260}{\sqrt{37}}$

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