Cities P,Q,R,S are such that Q is 4km due east of P,R is 3km due north of Q and S is 4km S 50degree W from R. Calculate the distance and bearing of P from S short and straight.

Answer 1

Using vectors:
PS = 4(cos0,sin0) + 3(cos90°,sin90°) + 4(cos220°,sin220°)
= (4,0) + (0,3) + (-3.0642, -2.5712)
= (.9358, .4288)

magnitude = √(.9358^2 + .4288^2) = appr 1.029
angle = tan^-1 (.4288/.9358) = 24.62°
or using your notation: N 65.38° E

Answer 2

All angles are measured CCW from +x-axis.
Given:
PQ = 4km[0o].
QR = 3km[90o].
RS = 4km[50o] W. of S. = 4km[220o] CCW.
SP = ?.

SP = 4[0o] + 3[90] + 4[220o].
X = 4*Cos0 + 3*Cos90 + 4*Cos220 = 4 + 0 - 3.06 = 0.94 km.
Y = 4*sin0 + 3*sin90 + 4*sin220 = 0 + 3 - 2.57 = 0.43 km.
SP = 0.94 + 0.43i = 1.03km[24.6o] CCW.