an airplane is flying a compass of 340 degrees at 325 mph. A wind is blowing with the bearing 300 degrees at 30 mph. Find the component form of the velocity, and find the actual ground speed and direction of the plane.

Answer 1

V = 325mph[340o] + 30mph[300o]

X = 325*cos340 + 30*cos300 = 320.4 mph =
Hor. component.
Y = 325*sin340 + 30*sin300 = -137.1 mph
= Ver. component.

TanA = Y/X = -137.1/320.4 = -42802
A = -23.17o,CW = 336.8 CCW.

V = 320.4/cos336.8 = 348.6mph[336.8o]

Answer 2

V= <325 cos110, 325 sin110>
W=<40cos130, 40sin130>
Resultant =V+W
Actual ground speed = magnitude of resultant = √(-136.87)^2+(336.04)^2 = 362.85 mph