math  bearings
posted by anonymous .
Been working on this question for a while. Have drawn a diagram (looks like a tilted trapezium, correct me if i'm wrong), but I still can't find any right angles to calculate the distance. Help would be appreciated, thanks.
Question:
A plane flies on a true bearing of 320° for 450 km. It then flies on a true bearing of 350° for 130 km and finally on a true bearing of 050° for 330 km. How far north of its starting point is the plane?

math  bearings 
Damon
First of all you do not fly on a "bearing". You fly, or sail, on a "heading". A "bearing" is the direction a lighthouse or whatever is from you. Math texts are totally ignorant of navigation.
Anyway, let's do it in conventional xy axes:
320 is 50 degrees above x axis
so
in x y coordinates
x1 = 450 cos 50
y1 = +450 sin 50
350 is 80 degrees above x axis
so
in x y coordinates
x2 = 130 cos 80
y2 = +130 sin 80
50 is 40 degrees above +x axis
so
in x y coordinates
x3 = 330 cos 40
y3 = 330 sin 40
NOW ADD those orthogonal (perpendicular to each other) vector components
x = x1 + x2 + x3
y = y1 + y2 + y3
north dist = y
no need to even calculate sqrt(x^2 + y^2) :) 
math  bearings 
anonymous
Wow, you are really good at this, and also BIG thanks for the extremely quick reply and easytounderstand solution <3

math  bearings 
Damon
You are welcome :)
( I have spent some time at sea :) 
math  bearings 
Henry
When using bearings, multiplying the distance by the Cos of the angle gives the vertical component instead of the usual hor. component. We need the vertical component only.
Displacement = 450km[320o] + 130[350o] + 330[50o].
Y = 450*Cos320 + 130*Cos350 + 330*Cos50 = 344.72 + 128.03 + 212.12 = 685 km, North.
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