physics
posted by marissa on .
A 316kg boat is sailing 12.0° north of east at a speed of 1.60 m/s. 39.0 s later, it is sailing 36.0° north of east at a speed of 4.00 m/s. During this time, three forces act on the boat: a 31.4N force directed 12.0° north of east (due to an auxiliary engine), a 20.7N force directed 12.0° south of west (resistance due to the water), and (due to the wind). Find the (a) the magnitude and (b) direction of the force . Express the direction as an angle with respect to due east.

Fn = 31.4N @ 12o,CCW + 12.7N @ 192o,CCW.
X=Hor.=31.4*cos12+12.7*cos192=18.3 N.
Y=Ver.=31.4*sin12+12.7*sin192=3.89 N.
a. Mag.=sqrt((18.3)^2+(3.89)^2=18.7 N
b. tanA = Y/X = 3.89/18.3 = 0.21257
A = 12o, CCW = 12o North of East.