[OSM-newbies] newbies Digest, Vol 27, Issue 27

[Mike Harris]

Oh dear! High school trigonometry at my advanced age! I only hope that I
have understood the question correctly. If so, the short answer is yes.

The simplest approach is to take each of points A, B and C separately as the
solution for A does not require information about B or C.

As I understand the question, you know the positions of X and Y - and
therefore you know the distance between them - call it P - and the bearing
of Y from X - call it L. You also know the bearing of unknown point A from
point X - call it J - and the bearing of unknown point A from point Y - call
it K.

Consider the triangle XYA. You know that the distance XY = P. You know the
angle at X (the difference between the bearings from X of points A and Y) =
J+L. Based on the simple properties of parallel lines it should be clear
that the angle at A (the difference between the bearings of A from X and
from Y) = K-J. As a triangle's internal angles sum to 360 degrees, you know
that the angle at Y is 360-(J+L)-(K-J) = 360-K-L.

Therefore in triangle XYA you know one side and two angles - so this
triangle is completely defined and can be solved. Solve it for the distance
XA - the distance from X to A. Call this distance Q. Then, applying the sine
rule:

P / sin (K-J) = Q / sin (360-K-L)

So Q = P * sin (360-K-L) / sin (K-J)

Now that you know Q, it is simple to calculate the distance of A north/south
from X ( = Q cos J) and the distance of A east/west from X ( = Q sin J).
Thus you have the northing and easting for point A.

Repeat for points B and C.

I hope that I have (a) got it right and (b) made it clear - it is over 50
years since I did this stuff in what would probably nowadays be called high
school.

What a way to spend a Bank Holiday afternoon!

Cheers!







Mike Harris

-----Original Message-----
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Sent: 25 May 2009 12:00
To: newbies at openstreetmap.org
Subject: newbies Digest, Vol 27, Issue 27

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