Question
An airplane has an airspeed of 770 kilometers per hour at a bearing of N 48 degrees E. If the wind velocity is 28 kilometers per hour from the west, find the angle representing the bearing for
the ground speed?
I found Rx and Ry which are 543.23 and 572.22 respectively. To find the angle I took tan^-1 of 572.22/543.23 and got 46. 489 for the angle, but the answer is wrong. R=789 and I got that right so I don't understand why the angle is incorrect.
the ground speed?
I found Rx and Ry which are 543.23 and 572.22 respectively. To find the angle I took tan^-1 of 572.22/543.23 and got 46. 489 for the angle, but the answer is wrong. R=789 and I got that right so I don't understand why the angle is incorrect.
Answers
Vp + 28[0o] = 770[90-48] = 770[42o]CCW
Vp + 28 = 770*cos42 + i770*sin42
Vp + 28 = 572.2 + 515.2i
Vp = 544.2 + 515.2i = 749mi/h[43.6o] CCW
= 749mi/h[46.4o] E of N.
Vp + 28 = 770*cos42 + i770*sin42
Vp + 28 = 572.2 + 515.2i
Vp = 544.2 + 515.2i = 749mi/h[43.6o] CCW
= 749mi/h[46.4o] E of N.