Three towns A,B&C are located such that the distance between A&B is 15km and the distance between A&C is 90km, if the bearing of B from A is 075 and the bearing of C from A is 310.

Question

(a) draw the diagram
(b) find the distance between B and C
(c) find the bearing of C from B

Answer 1

If A is at origin
Then C is 360-310 = 50 degrees west of north
and B is 75 degrees east of north
total angle CAB = 50 + 75 = 125 degrees
law of cosines
a^2 = b^2 + c^2- 2 b c cos 125
a^2 = 90^2 + 15^2 - 2 (90)(15)cos 125
solve for a
then use law of sines for other angles
if you find angle B for example you know B +75 + angle of BC to north south line is 180

Answer 2

I will assume you made the sketch
then you can see that the angle CAB = 125°
and we have a clear case of the cosine law.
BC^2 = 90^2 + 15^2 - 2(90)(15)cos 125°
= ....
Find BC
Once you have BC, use the sine law to find one of
the other angles. I suggest finding angle C to avoid any
conflicts with the ambiguous case.
Of course once you have 2 angles of the triangle, the third is routine.

Put all your angles in your sketch and you should be able to find the
bearing of C from B.

Answer 3

All angles are measured CW from +y-axis
b. BC = AC - AB = 90[310o] - 15[75o],
X = 90*sin310 - 15*sin75 = -83.4 km..
Y = 90*Cos310 - 15*Cos75 = 50.1 km.
BC = -83.4 + 50.1i = 97.3 km[-59o] = 97.3km[59o] W. of N. = 97.3km[301o].

c. Bearing from B to C = 301o(see part b).