Let vector u point from A to B, and let v point from B to C. Then
u = (-50 sin30°, -50 cos30°) = (-25.00, -43.30)
v = (80 sin30°, -80 cos30°) = (40, -69.28)
Use law of cosines to find AC:
AC^2 = 50^2 + 80^2 - 2*50*80 cos120°
The vector from A to C is u+v = (15,-112.58)
bearing of A from C = 90-θ where tanθ = -15/112.58
eastward from B to C is just 40
An+aeroplane+flies+from+a+town+A+to+a+town+B++50km+away+on+a+bearing+of+210+degree+.if+then+flew+from+town+B+to+another+town+C+on+a+bearing+of+150degree.if town C is 80km away from B,find the distance between A and C to the nearest km.find the bearing A from C to the nearest degree.find how far east B is C.
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