Proof Tri. Equstion?

Let P be the point that is on line 1(N1, E1) 2(N2,E2) and D1 from 1. So the distance of P from N ( the y - axis) = D1 sin B and the distance of P from E ( the x -axis) = D1 cos B.
Let PQ be the line perpendicular from P to the x -axis, so distance of Point (N3,E3) to PQ = D2 cos B and the vertical distance of P above (N3,E3)
= D2 sin B.
So we get
N3 - N1 = D1 cos B - D2 sin B..................(1)
E3 - E3 = D1 sin B + D2 cos B...................(2)
(1) x sin B
(N3 - N1) sin B = D1 cos B sin B - D2 sin ^2 B.............(3)
(2) x cos B
(E3 - E1) cos B = D1 cos B sin B + D2 cos ^2 B ............(4)
(4) - (3) we get
(E3 - E1) cos B - (N3 - N1) sin B = D2 ( cos ^2 B + sin ^2 B)
so D2 = (E3 - E1) cos B - (N3 - N1) sin B
Sub this result into either (3) or (4), you will find D1.




2009-04-22 17:09:35 補充：
Correction: Equation (2) should be E3 - E1, not E3 - E3.

2009-04-22 17:13:34 補充：
To find D1, you may multiply (1) by cos B and multiply (2) by sin B and add the 2 equations together.

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