Question

A village r is 10km from a point p on a bearing 025degree from p. Another village A is 6km from p on a bearing 162degree.calculate distance of r from a and the bearing of r from a.

Answer 1

From my diagram, I have triangle PRA with angle P at 137°, PR = 10 and PA = 6
AR^2 = 6^2 + 10^2 - 2(6)(10)cos137°
I get AR = 14.96 km

sinR/6 = sin137/14.96
R = 15.875°
bearing = 25-15.875 = appr 9.1°

or by vectors:
vector AR = (10cos25,10sin25) - (6cos162,6sin162)
= (14.769... , 2.372..)
magnitude = √(14.769..^2 + 2.372..^2) = 14.956 , same as using the cosine law.
tanØ = 2.372/14.769
Ø = 9.1°

Answer 2

All angles are measured CW from +y-axis.
AR = AP+PR = 6[162+180] + 10[25o].
AR = (6*sin342+10*sin25) + (6*cos342+10*cos25)i
AR = 2.37 + 14.8i = 15km[9.1o].