# Distance and Azimuth Calculation

I hope you could help me on this.
Is there anyone can tell how to calculate azimuth and distance of two points on earth, the points is in two-dimensional cartesian format, using spherical trigonometri for more accurate calculation. If you have reference or you have sample of answered exercise how to calculate it, please let me know. Thx.

are you refering to aeronautical navigation?

no, what I mean is how to calculate distance and azimuth/bearing if you have a pair of site coodinates (in two-dimensional cartesian format) using spherical trigonometri (spherical triangels, not just a planar triangels) ?

you can use

x = p sin @ cos #

y = p sin @ sin #

z = p cos @

where p is the distance from the center of the sphere

@ is the angle made from the positive x-axis to the pprojection of the point onto the x y plane

and # is the angle from the positive z - axis to the point

All this is calculated by the sinus and cosine rule. In any spherical triangle call the corners A, B and C and the opposite sides a, b and c. Then the sinus rule is: sin(A)/sin(a)=sin(B)/sin(b)=sin(C)/sin(c) or in words: the ratio between the sinus of an angle and the sinus of its opposite side is equal for all three pairs.
The cosine rule: cos(a)=cos(b)cos(c)+sin(b)sin(c)cosA which may be applied for the other sides as well of course.
So in calculating distance and azimuth (or bearing/course) let distance be d, the latitude of place A is a and of place B is b and the angle between the meridians at the pole is the difference in longitude, which I'll call P, then applying the cosine rule gives you:
cos(d)=cos(90-a)cos(90-b)+sin(90-a)sin(90-b)cos(P) or
cos(d)=sin(a)sin(b)+cos(a)cos(b)cos(P)
Then calculate azimuth by the sinus rule:
sin(A)=sin(P)sin(90-b)/sin(d) from A to B or sin(B)=sin(P)sin(90-a)/sin(d) from B to A.
or sin(A)=sin(P)cod(b)/sin(d) and sin(B)=sin(P)cos(a)/sin(d) resp.
Please remember to "make sense" of your output, because for example invsin(0,5) gives you 030, 150, 210 and 330.