Two ships leave a harbor at the same time, traveling on courses that have an angle of 140 degrees between them. If the first ship travels at 26 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 3 hours?

Answer 1

I drew the vector for ship # 1 @ 0 deg
and ship # 2 @ 140 deg.

d1 = 26 mi / h * 3 h = 78 mi @ 0 deg.

d2 = 34 mi / h * 3 h = 102 mi @ 140 deg.

d = 102 mi @ 140 deg - 78 mi @ 0 deg.

X = hor = 102 * cos140 - 78cos0,
= -78.1 - 78 = - 156.1 mi.

Y = ver = 102sin140 - 78sin0
= 65.6 - 0 = 65.6 mi.

d^2 = x^2 + y^2,
= 24367.2 + 4298.7,
= 28665.9,
d = sqrt(28665.9) = 169.3 miles apart.