If `vec v_"XY"` represents the velocity of X with respect to Y, then

`vec v_"AB" + vec v_"BC" = vec v_"AC"`.

In other words, we can eliminate the B and go straight from A to C, provided we know the velocity of B relative to C. Sometimes you will have to rearrange this equation to solve for a different velocity—that’s about as complicated as these problems get.

An airplane has a heading of 245 km/h [N 17º E], and the wind is coming from [S 35º W] at 89.0 km/h. What is the course?

First, we need to familiarize ourselves with airplane terminology:

- heading (`vec v_"PA"`)
- velocity of the plane with respect to the air
- wind (`vec v_"AE"`)
- velocity of the air with respect to the Earth
- course/groundspeed (`vec v_"PE"`)
- velocity of the plane with respect to the Earth

The velocity of the wind is a vector pointing in the direction in which it is going. Since the wind is coming *from* [S 35º W], it is going *to* [N 35º E].

Since `vec v_"PE" = vec v_"PA" + vec v_"AE"`, the course is

`vec v_"PE" = 245\ "km/h [N 17º E]" + 89.0\ "km/h [N 35º E]"`

`= 331\ "km/h [N 22º E]"`.

I didn’t show my steps here. You need to draw a sketch and make a table. Refer to the vectors section if you don’t know how to add vectors.