The bearing of B from A 12km is 038°.the distance between A and C is 17km.if C is on a bearing 152° from A find:(a) the bearing of A from C (b) the distance between point B and C (C) the bearing of C from B

Answer 1

(a) add 180°
(b) in triangle ABC, with the usual labeling of sides and angles, angle A = 114°, so using the law of cosines, that distance is
a^2 = 12^2 + 17^2 - 2*12*17 cos114°
(c) use the law of sines to find angle C:
sinC/c = ainA/a
Now you can use C to find the bearing.

Answer 2

All angles are measured CW from +y-axis.
Given: AB = 12km[38o], AC = 17km[152o]. BA = 12km[38+180].
a. CA: bearing(direction) = 152+180 = 332 deg.

b. BC = BA [218]+AC[152] = 12[218']+17[152]
BC = (12*sin218+17*sin152)+(12*cos218+17*cos152)i
BC = 0.593-24.5i.
BC = sqrt(0.593^2+24.5^2) = 24.51km[-1.4o] = 24.51km[1.4o] W. of N.

c. Bearing(direction) = 1.4 deg W. of N. = 358.6 deg CW.