Question

an airplane flies on a compass heading of 90.0 degree at 320 mph. the wind affecting the plane is blowing from 312 at 37.0mph. what is the true course and ground speed of the airplane..

Answer 1

vector addition

x direction is east, so plane airspeed is (320,0) no y (north) component

wind from 317o pushes toward 47o south of east
has x contribution to velocity of 37 cos(47) = 25.2
has y contribution to velocity of -37sin(47) = -27.1
so wind vector to add to plane is (25.2, -27.1)

components add to get resultant (345.2, -27.1)

speed is square root of sum of these components squared,

s = 346.3 mph

direction is angle whose tangent is (-27.1/345.2) or 4.5o south of east

thus 94.5o

Low wind, small velocity, little change in direction and speed.

You should check my math, though.

Answer 2

it says it is blowing from

Answer 3

< 320 cos 0 , 320 sin 0> + <37 cos 312, 37 sin 312>
= < 320 cos 0 + 37 cos 312 , 320 sin 0 + 37 sin 312>
magnitude of resultant vector
= sqrt ( (320 cos 0 + 37 cos 312)^2 + (320 sin 0 + 37 sin 312)^2)
direction

Answer 4

Yes, wind toward 48o south of east, not 47o south of east, my error. Correct outcome slightly different from mine, as done right by @perl. Thanks.

Answer 5

im afraid this whole problem is wrong