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P,Q and R are points on the same horizontal plane .the bearing of Q from P is 150° and the bearing of R from Q is 060°. if |PQ| =5m and |QR|=3m find the bearing of R from P correct your answer to the nearest degree

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  1. In geometry, drawing a sketch from given information is half-way to the solution.

    Note bearing is measured clockwise from North.

    If you draw the sketch in Cartesian coordinates, using P at the origin, you will have
    P(0,0)
    Bearing 150°≡ θ=90-150=-60° in Cartesian coordinates.
    So
    Q(5cos(-60), 5 sin(-60))
    =Q(2.5, -2.5&radic/3)

    Similarly, the bearing of R from Q is 60°≡ θ=90-60=30°.
    So
    Δx=3cos(30)=1.5√3
    Δy=3sin(30)=1.5

    Therefore the coordinates of R relative to P
    R(2.5+Δx, -2.5&radic/3+Δy)
    =R(2.5+1.5√3, -2.5√3+1.5)

    Bearing of R from P
    =90-atan(y/x)
    =90-atan((-2.5√3+1.5)/(2.5+1.5√3))
    =90-(-29)
    =119°

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