# Plane bearing

1. Apr 30, 2006

### James_fl

[REVISED]My apology for forgetting to post the question..

Hello, I have a problem to solve this question. I am sure that I have done a mistake somewhere since it is impossible to solve: 246.25 sin x - 676.58 cos x = -27768.42 (last line)

Question:

Use Cartesian vector method to solve all the problems. If you use any other method, you will receive zero.
A pilot wishes to fly form city A to city B, a distance of 720 km on a bearing of 070 degree. The speed of the plane is 700 km/h. An 60 km/h wind is blowing on a bearing of 110 degree. What heading should the pilot take to reach his or her destination? How long will the trip take?

This is my work:

http://i66.photobucket.com/albums/h242/jferdina/Bearing.jpg" [Broken]
http://i66.photobucket.com/albums/h242/jferdina/Bearing-continued.jpg" [Broken]

Thank you..

James

Last edited by a moderator: May 2, 2017
2. Apr 30, 2006

### HallsofIvy

I would interpret "use Cartesian vector method" as meaning write the vectors in terms of x and y components. in particular, "60 km/h on a bearing of 110 degrees" will have x and y components 60 cos(110) and 60 sin(110) respectively: the wind velocity vector is 60 cos(110)i+ 60 sin(110)j. Suppose the velocity vector of the airplane (relative to the air) is ui+ vj so that the airspeed of the airplane is $\sqrt{u^2+ v^2}= 700$. Then the velocity vector relative to the ground is ui+ vj+ 60 cos(110)i+ 60 sin(110)j= (u+ 60 cos(110))i+ (v+ 60 sin(110))j.

The position vector of city B, relative to city A, is 720 cos(70)i+ 720 sin(70)j. To go from city A to city B, in t hours, we must have
(u+ 60 cos(110))ti+ (v+ 60 sin(110))tj= 720 cos(70)i+ 720 sin(70)j.
That, together with $\sqrt{u^2+ v^2}= 700$ gives 3 equations to be solved for the 3 variables u, v, and t.