Question

Two ships leave a harbor at the same time. One ship travels on a bearing s11w at 16 miles per hour. The other ship travels on a bearing n15e at 12 miles per hour. How far apart will the ships be after 2 ​hours?

Answer 1

I suggest you do this in x y coordinates, x is East and y is North
16 knots for 2 hours is 32 nautical miles
a heading (not a bearing by the way) of South 11 West is 11 degrees clockwise beyond the -y axis
so x location = -32 sin 11 = -6.10
and
y location = - 32 cos 11 = - 31.4
so first ship is at (-6.10 , -31.4 )
The second ship travels N15 E at 12 knots for 2 hours
that is 90 - 15 = 75 deg above x axis
distance = 24 nautical miles
x = 24 cos 75 = 6.21
y = 24 sin 75 = 23.2
so location is (6.21 , 23.2)
now distance between
d^2 = x difference^2 + y difference^2
=
d ^2 = (6.21 + 6.1)^2 + (23.2 + 31.4)^2
I think you can find d now.

Answer 2

the angle between the two courses is 176°
The distance d is thus
d^2 = 32^2 + 24^2 - 2*32*24 cos176°

a very good approximation is just 32+24 = 56 miles, since the two ships are traveling in almost opposite directions.