Question

One ship leaves a port at 1 pm traveling at 12 mph at a heading of 70 degrees . at 2 pm another ship leaves the port traveling at 15 mph at a heading of 120 degrees . find the distance between the ships at 4 pm.

Answer 1

28.4 miles.
   This problem effectively creates a triangle with the data you're giving being SAS (side, angle, side) and then asks you for the length of the 3rd side. So let's first determine the sides and angle of the triangle.
 1st side, the ship leaves at 1 pm and travels for 4 - 1 = 3 hours at 12 mph. So the length of the side is 12 * 3 = 36 miles.
 2nd side, the ship leaves at 2 pm and travels for 4-2 = 2 hours at 15 mph. So the length of the side is 15 * 2 = 30 miles
 The included angle. The key thing to realize is that the angle isn't either of the ship headings. It's the difference between the headings. So we have one ship with a heading of 70 degrees and the other ship with a heading of 120 degrees. So there's a difference of 120 - 70 = 50 degrees between the two ships.
So we have side 1 = 36 miles, side 2 = 30 miles, included angle = 50 degrees.
 Since we we only know the lengths of the adjacent sides to the known angle, we should use the law of cosines to solve for the unknown side (if we had a known angle and knew the length of the opposite side to the angle, we would use the law of sines).
The law of cosines is:
 c^2 = a^2 + b^2 - 2ab cos θ
 where a,b = adjacent sides to known angle
 c = side opposite to known angle
 Î¸ = known angle
   Let's substitute the known values and calculate
 c^2 = a^2 + b^2 - 2ab cos θ
 c^2 = 36^2 + 30^2 - 2*36*30 cos 50
c^2 = 1296 + 900 - 2160*0.64278761
 c^2 = 1296 + 900 - 1388.421237
 c^2 = 807.5787631
 c = 28.41793031
    So at 4 pm, the distance between the ships is 28.4 miles.