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When doing this type of problem always start with a diagram.
In nil wind conditions, the ETP (CP) will be half-way between A and B. If a wind is blowing the ETP will be displaced upwind from the track mid-point position. Use this to check if you get a sensible answer after the calculations are done.
Track Home = 115 M
TAS = 140 kts
W/V = 250 M/30
GShome = 160 kts
DETP = (Total Dist x GShome)/(2xTAS) = (320 x 160)/(2 x 140) = 182.8 = 183 nm
Note: 183 nm is more than half-way to B - this is to be expected as we are punching into a headwind on route to B.
Track Out = 295 M
TAS = 140 kt
W/V = 250 M/30
GSout = 117 kts
Some of the processes we use when working the wind triangle depend, for their validity, on the wind speed's being a sensibly small fraction of the TAS. As that fraction increases, things can require a little more care and precision and, in the extreme, the solution doesn't always function well at all.
For this particular case, the wind is 30 kt, a decent proportion of the TAS. The 2 x TAS approximation for (G/S on + G/S home) is starting to unravel a little.
Suggest you rework the calculations to check out the difference. Just for interest ....
Engineering specialist in aircraft performance and weight control.