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 :)

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 
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) + (3.125i5.41) = 3.46 + 1.14i = 3.64km[18.2o] = 3.64km[288.2o].
Direction: 3.64km at a bearing of 288o. 
Henry
Correction: Tan A = X/Y = (3.46)/1.14 = 3.03509.
A = 72o = 72o N. of W. = 342o Clockwise from +yaxis
Direction: 3.64km at a bearing of 342o.
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