Two ships A and B are sailing straight away from a fixed point O along routes such that $$\angle AOB$$ is always 120°. 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):

We are given two ships A and B moving away from a fixed point O such that the angle ∠AOB is always 120°. At a specific moment, OA = 8 km, OB = 6 km, ship A is moving away from O at 20 km/hr, and ship B at 30 km/hr. We need to find the rate at which the distance between A and B is changing.

Let OA = x km, OB = y km, and the distance AB = z km. At the given instant, x = 8, y = 6, dx/dt = 20 km/hr, and dy/dt = 30 km/hr. The angle between OA and OB is constant at 120°.

Using the law of cosines in triangle OAB:

$$ z^2 = x^2 + y^2 - 2xy \cos(120^\circ) $$

Since cos(120°) = -1/2, substitute:

$$ z^2 = x^2 + y^2 - 2xy \left(-\frac{1}{2}\right) $$

Simplify:

$$ z^2 = x^2 + y^2 + xy $$

Now, differentiate both sides with respect to time t. Differentiate implicitly:

$$ \frac{d}{dt}(z^2) = \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) + \frac{d}{dt}(xy) $$

Apply the chain rule and product rule:

$$ 2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} + \left( x \frac{dy}{dt} + y \frac{dx}{dt} \right) $$

Simplify:

$$ 2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} + x \frac{dy}{dt} + y \frac{dx}{dt} $$

Group terms:

$$ 2z \frac{dz}{dt} = (2x + y) \frac{dx}{dt} + (2y + x) \frac{dy}{dt} $$

At the given instant, x = 8, y = 6. First, compute z:

$$ z^2 = (8)^2 + (6)^2 + (8)(6) = 64 + 36 + 48 = 148 $$

So,

$$ z = \sqrt{148} = \sqrt{4 \times 37} = 2\sqrt{37} $$

Now, substitute the values into the differentiated equation:

$$ 2 \times (2\sqrt{37}) \frac{dz}{dt} = (2 \times 8 + 6) \times 20 + (2 \times 6 + 8) \times 30 $$

Compute the coefficients:

$$ 2x + y = 2 \times 8 + 6 = 16 + 6 = 22 $$

$$ 2y + x = 2 \times 6 + 8 = 12 + 8 = 20 $$

So,

$$ 4\sqrt{37} \frac{dz}{dt} = 22 \times 20 + 20 \times 30 $$

Calculate the right-hand side:

$$ 22 \times 20 = 440 $$

$$ 20 \times 30 = 600 $$

$$ 440 + 600 = 1040 $$

Thus,

$$ 4\sqrt{37} \frac{dz}{dt} = 1040 $$

Solve for dz/dt:

$$ \frac{dz}{dt} = \frac{1040}{4\sqrt{37}} = \frac{1040}{4} \times \frac{1}{\sqrt{37}} = 260 \times \frac{1}{\sqrt{37}} = \frac{260}{\sqrt{37}} $$

Therefore, the distance between A and B is changing at the rate of $$\frac{260}{\sqrt{37}}$$ km/hr.

Comparing with the options, Option A matches.

Hence, the correct answer is Option A.