Posted by Anonymous on Wednesday, August 31, 2011 at 2:07am.
a. 32 deg North of West = 148 deg.,CCW.
X=hor. = -200 + 350cos148 = -496.82km.
Y = ver. = 350sin148 = 185.47km.
d = sqrt(X^2+Y^2),
d = sqrt((-496.82)^2 + (185.47)^2 =
530.3km.
b. tanA = Y / X = 185.47 / -496.82 = -0.37332.
A = -20.47 deg,CW. = 180 - 20.47 = 159.53 deg.,CCW. = 20.47 deg, North of West..
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