A nuclear submarine leaves its base and travels at 23.5 mph. For 2.00 hrs it travels along a course 32.1 degrees north of west. It then turns an additional 21.5 degrees north of west and travels for another 1.00 hrs. How far from its base is it?

Answer 1

d1 = 23.5mi/h * 2h = 47 Mi[180o-32.1o] =
47Mi[147.9o]

d2 = 23.5mi/h * 1h = 23.5mi[180-53.6] =
23.5mi[126.4o]

d1 + d2 = 47mi[147.9] + 23.5mi[126.4o]
X = 47*cos147.9 + 23.5*cos126.4=-53.8 mi
Y = 47*sin147.9 + 23.5*sin126.4=43.9 mi

D^2=X^2 + Y^2=-53.8^2 + 43.9^2=4821.65
D = 69.4 miles.