# physics

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33km @ 90+10 = 33km @ 100 deg.,CCW.
18km @ 270+45 = 18km @ 315 deg.,CCw.
14km @ 270+50 = 14km @ 320 deg.,CCW.

X=hor.=33cos100 + 18cos315 + 14cos320 = 5.73 + 12.73 + 10.72 = 17.7km.

Y=ver.=33sin100 + 18sin315 + 14sin320
= 2.95 + (-12.73) + 9.0 = -0.78km.

TanA = Y / X = -0.78 / 17.7 = -0.04407
A = -2.52 deg = 2.52 deg South of East.

D = -0.78 / sin(-2.52) = 17.67km @ 2.52
deg South of East.

The return path is 17.67km @ 2.52 deg
North of East.

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