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
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 =
All angles are measured CW from +y-axis.
To find the bearing of A from B:
1. Start with the bearing of A, which is 141°.
2. The man continues his trip from town B on a bearing of 255°, which means he turns 255° from north towards the west.
3. To find the bearing of A from B, subtract the bearing of A from the bearing of B:
Bearing of A from B = Bearing of B - Bearing of A
= 255° - 141°
= 114°
Therefore, the bearing of A from B is 114°.
To find the bearing of A from C:
1. We are given the angle BCA as 35°.
2. The bearing of A from C can be found by subtracting the angle BCA from the bearing of A from B.
Bearing of A from C = Bearing of A from B - Angle BCA
= 114° - 35°
= 79°
Therefore, the bearing of A from C is 79°.
To find the bearing of A from B, you can subtract the bearing of A from the bearing of B.
The bearing of A from B is given as 255.
The bearing of A from B = 255 - 180 = 75.
Therefore, the bearing of A from B is 75.
To find the bearing of A from C, you can subtract the bearing of C from the bearing of A.
The angle BCA is given as 035.
The bearing of C from A = 180 + 35 = 215.
Therefore, the bearing of A from C is 215.