trg

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A plane flies in a direction of N70°E for 80 km and then on a bearing of S10°W for
150 km.
a How far is the plane from its starting point?
b What direction is the plane from its starting point?

  • trg -

    I will convert the directions to vectors using conventional angles
    first leg: (80cos20° , 80sin20°)
    2nd leg: (150cos260° , 150sin260°)

    resultant
    = (80cos20° , 80sin20°) + (150cos260° , 150sin260°)
    = (49.128, -120.3596)

    magnitude = √(49.128^2 + (-120.3596)^2) = 130 km
    direction:
    tanØ = -120.3596/49.128
    Ø = -67.8° or 292.2°
    or
    S 22.2° E

  • trg -

    or

    Make a sketch , let the original position be A, end of first leg B, and C the final position.
    I have a triangle ABC with angle B = 60°
    AB = 80 and BC = 150
    by the cosine law:
    AC^2 = 80^2 + 150^2 - 2(80)(150)cos60
    = 16900
    AC = √16900 = 130

    by sine law:
    sinA/150 = sin60/130
    sinA = 150sin60/130 = .99926
    angle A = 87.8°
    looking at your diagram, subtracting 20° yields the same angle of 67.8° as in my first solution.

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