Help! Question Below..?

The route followed by a hiker consists of three displacement vectors A, B and C. Vector A is along a measured trail and is 3400 m in a direction 24.1° north of east. Vector B is not along a measured trail, but the hiker uses a compass and knows that the direction is 40.6° east of south. Similarly, the direction of vector C is 33.7° north of west. The hiker ends up back where she started, so the resultant displacement is zero, or A + B + C = 0. Calculate the magnitude of vector B.

Help! Question Below..?
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Taking west to east as x- axis and south to north as y - axis





Resolving,vectors A(of magnitude 3400),B and C into components and equating the components along positive x - axis and negative x - axis





B*sin40.6=C* cos33.7-3400*cos24.1





B*0.65077 =C*0.83195 - 3400*0.91283





Dividing by 0.65077 onboth sides,





B =C*1.27841 - 4769.1535..............(1)





Equating the components along positive y - axis and negative y - axis





B*cos40.6 = 3400*sin24.1 + C *sin 33.7





B0.7593 = 3400*0.40833 + C*0.55484..........(2)





Substituting B from equation(1) in equation (2)





(C*1.27841 - 4769.1535)0.7593 = 1388.322 + C*0.55484





C*0.9707 - 3621.2183 = 1388.322 + C*0.55484





C*0.4159 =5009.5403





C = 12045.06 m





substituting for C in eq (1),





B =12045.06 *1.27841 - 4769.1535





B = 15398.525 - 4769.1535





B = 10629.37165 m


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Magnitudes of vectors A,B and C are given below





A = 3400 m





B = 10629.37165 m





C = 12045.06 m


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