Moving vertex along bisector to set a specific angle

Question

I've got a three nodes A, B, C which create an angle α.

Question: how to find a position of B' (x,y coordinates) which is effectively a shift of B along α bisector, that AB'C angle is equal to α+5°

Answer 1

Here (with bisector) simpler solution is possible.

Angle BCB' is 2.5 degrees

Side BB' of triangle BCB' might be calculated using Law of Sines as

Len(BB') = Len(BC) *  sin(2.5) / sin(alpha/2+2.5) 

Unit bisector vector

bis = BC/Len(BC) + BA/Len(BA)
bis = bis / Len(bis)

B' coordinates

B' = B + bis * Len(BB')