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.

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  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°

  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].

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