A man travels from town A on the bearing of 141,he then decide to continue with his trip from town B on a bearing of 255 to the angle BCA=035.find the bearing of A from b and A from c

Answer 1

Disp. = AB[141] + BC[255o].
Use unit vectors as magnitude:
Disp. = 1[141o] + 1[255o].
X = 1*sin141 + 1*sin255 = -0.337.
Y = 1*Cos141 + 1*Cos255 = -1.036.
Disp. = X + Yi = -0.337 - 1.036i = 1.09[18o].

Bearing of BA = 141 + 180 =

Bearing of AC = 18o.
Bearing of CA = 18 + 180 =

Answer 2

All angles are measured CW from +y-axis.