Asked by MARY
3 town abc are such that the distance a and b is 50km and distance between ac is 90km if the bearing of b from a is 075 and the bearing c from a is 310 find the distance bc b bearing of c from b
Answers
Answered by
oobleck
Draw a diagram. In triangle ABC, ∠A = 125°
So, using the law of cosines to find a (BC), we have
a^2 = 50^2 + 90^2 - 2*50*90 cos125°
BC = a = 125.55
The locations of the towns are
A = (0,0)
B = (48.3,12.94)
C = (-68.94,57.85)
So, measuring counterclockwise from the +x direction, we have
tanθ = (57.85-12.94)/(48.3+68.94) = -0.383
θ = 180-21 = 159°
So the bearing of C from B is 270+21 = 291°
So, using the law of cosines to find a (BC), we have
a^2 = 50^2 + 90^2 - 2*50*90 cos125°
BC = a = 125.55
The locations of the towns are
A = (0,0)
B = (48.3,12.94)
C = (-68.94,57.85)
So, measuring counterclockwise from the +x direction, we have
tanθ = (57.85-12.94)/(48.3+68.94) = -0.383
θ = 180-21 = 159°
So the bearing of C from B is 270+21 = 291°
Answered by
Fawwaz
Thanks you have been a good help
Answered by
Akor micheal
three towns a and c from the distance between a and b From a is 50km and the distance between a and c is 90km if the beaming of b is 75 and the bearing of c from a is 310 find
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