I will convert the directions to vectors using conventional angles
first leg: (80cos20° , 80sin20°)
2nd leg: (150cos260° , 150sin260°)
= (80cos20° , 80sin20°) + (150cos260° , 150sin260°)
= (49.128, -120.3596)
magnitude = √(49.128^2 + (-120.3596)^2) = 130 km
tanØ = -120.3596/49.128
Ø = -67.8° or 292.2°
S 22.2° E
Make a sketch , let the original position be A, end of first leg B, and C the final position.
I have a triangle ABC with angle B = 60°
AB = 80 and BC = 150
by the cosine law:
AC^2 = 80^2 + 150^2 - 2(80)(150)cos60
AC = √16900 = 130
by sine law:
sinA/150 = sin60/130
sinA = 150sin60/130 = .99926
angle A = 87.8°
looking at your diagram, subtracting 20° yields the same angle of 67.8° as in my first solution.
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