Asked by Neha
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
Answered by
Henry
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.