If there is unknown current (speed and direction) and leeway angle is unknown too, then the answer is no. But TWA, TWS and AWA already give you AWS if there is no leeway and no current.
Are you 100% sure?
I stated there is enough information there, not that all is needed for solution.
TWA and AWA relative to boat HDG are two angles of the wind triangle, so third is (trivially) known, yes?
BTW what is that third angle called, I don't know the word in English?
We have one length/side (TWS) of the triangle, so the two others are (trivially) known, yes?
Other sides being the AWS and SPD, yes?
Did I miss something? How can current and leeway skew a rigid triangle?
TWD is not useful at all, and SOG is only relevant for the case current and leeway are to be accounted for, and is not enough if both factors are fully unknown.
TWD and SOG can then be used to solve the remaining unknown vectors, yes, like HDG etc?
Maybe I need to draw this on a sketchpad, I did not think it through so thoroughly...