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Problem

Two ships leave a port: Ship 1 travels due north for $7$ nautical miles, and Ship 2 travels $60^\circ$ east of north for $9$ nautical miles. Find the distance between the ships and the area of the triangle formed by the two ships and the port.

Step 1: Use the law of cosines for the ship-to-ship distance

The two travel paths form a triangle with sides $7$ and $9$ and included angle $60^\circ$. Using the law of cosines,

$$c^2 = 7^2 + 9^2 - 2(7)(9)\cos 60^\circ.$$

Since $\cos 60^\circ = \dfrac{1}{2}$,

$$c^2 = 49 + 81 - 63 = 67.$$

So the distance between the ships is

$$c = \sqrt{67} \approx 8.19.$$

Step 2: Use the area formula for the triangle

With two sides and the included angle, the area is

$$A = \frac{1}{2}(7)(9)\sin 60^\circ.$$

Because $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$,

$$A = \frac{63\sqrt{3}}{4} \approx 27.28.$$

Answer

The ships are $\sqrt{67} \approx 8.19$ nautical miles apart, and the triangle's area is $\dfrac{63\sqrt{3}}{4} \approx 27.28$ square nautical miles.