Asked by Mumbi
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
Answers
Answered by
henry2,
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 =
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 =
Answered by
henry2,
All angles are measured CW from +y-axis.
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