math - bearings

posted by anonymous

Starting from their base in the national park, a group of bushwalkers travel 1.5 km at a true bearing of
030°, then 3.5 km at a true bearing of 160°, and then 6.25 km at a true bearing of 300°. How far, and at what true bearing, should the group walk to return to its base?

Pretty straightforward question, but it gives a messy diagram, and so, I can't make anything out of it :L

Also big thanks to Damon for helping me out with all these bearing questions :)

  1. Damon

    first leg
    north +1.5 cos 30
    east +1.5 sin 30
    second leg
    north -3.5 cos 20
    east +3.5 sin 20
    third leg
    north +6.25 sin 30
    east -6.25 cos 30
    so where are they from start?
    north 1.5 cos 30 - 3.5 cos 20 + 6.25 sin 30
    east +1.5 sin 30 +3.5 sin 20 - 6.25 cos 30

    d = Total distance = sqrt(north^2+east^2)

    A = angle clockwise from north (compass true angle, not usual math angle)

    tan A = east/north

    pick the right quadrant by the signs of north and east distances

    to get back to origin go distance d at A + 180 if A <180 or at A - 180 if A>180

  2. Henry

    When using Bearings, multiplying the distance by the Cos of the angle gives the vertical component instead of the usual hor. component.

    Displacement = 1.5km[30o] + 3.5[160] + 6.25[300].
    Disp. = (1.3i+0.75) + (-3.29i+1.2) + (3.125i-5.41) = -3.46 + 1.14i = 3.64km[-18.2o] = 3.64km[288.2o].

    Direction: 3.64km at a bearing of 288o.

  3. Henry

    Correction: Tan A = X/Y = (-3.46)/1.14 = -3.03509.
    A = -72o = 72o N. of W. = 342o Clockwise from +y-axis
    Direction: 3.64km at a bearing of 342o.

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