Answer:
412 km/h S27°E
Step-by-step explanation:
You want the resultant ground speed of an airplane traveling 360 km/h at S40°E in a wind of 100 km/h toward S25°W.
Vectors
The direction vectors here are given relative to south. Ordinarily, bearing vectors would have their angles measured clockwise from north. (Those are the angles used in the second attachment.)
If we measure the angle CCW from south, then the direction vectors for plane and wind in (south, east) coordinates are ...
Plane = 360∠40° = 360(cos(40°), sin(40°)) ≈ (275.78, 231.4)
Wind = 100∠-25° = 100(cos(-25°), sin(-25°)) ≈ (90.63, -42.26)
Resultant
The resultant vector is the sum of these:
(275.78 +90.63, 231.40 -42.26) = (366.41, 189.14)
The speed of the plane is then ...
|r| = √(366.41² +189.14²) ≈ 412.35 ≈ 412 . . . . km/h
The direction of the plane is ...
∠ = arctan(189.14/366.41) ≈ 27.30° ≈ 27° . . . . . . east of south
The plane's ground speed is about 412 km/h at S27°E.
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Additional comment
We have used 3 different coordinate systems here.
The diagram in the first attachment shows the resultant angle measured CCW from east (297°). The second attachment shows calculator results using bearing angles measured CW from north. The rectangular coordinates in that calculation are (north + east·i) and the bearing angle is 153°. Above, we have used (south, east) coordinates with angles measured CCW (east) from south. It's not all that hard to keep it straight, as long as any given calculation uses consistent coordinate definitions.
297° CCW from +x is 27° CCW from -y and 153° CW from +y.
