Bearings are in relation to north. In the triangle ABC, angle A is the starting-point, AB=80km the first leg of the journey, and BC=120km, the last leg. Angle B=42+180-140=82 degrees.
Using the cosine rule,
AC^2=AB^2+BC^2-2AB.BCcosB=6400+14400-2*9600cos82=18127.88 approx.
Therefore AC=134.64km approx.
We can find other angles in ABC using the sine rule, so sinA/120=sin82/134.64,
and sinA=120*sin82/134.64=0.8826 approx.
From this A=61.96º. To find the bearing we need to add 42º, the initial bearing of B from A=103.96.
See diagram in similar recent question where an aircraft flew between three points P, Q and R.