Posted by Mike on .
Please can someone solve this PZX triangle for me and give your workings.
Angle ZPX 040 degrees
Distance PZ 3000M (NM)
Distance PX 2000M (NM)
Thanks a million.
Mike

Nautical/Maths 
drwls,
I don't see why you have both M and (NM) following the length numbers. Are the dimensions meters, nanometers, miles of nautical miles? Your call.
After deciding what units you are talking about, let side PZ of the triangle be a and side PX be b.
The length of side ZX (c) can be obtained by using the law of cosines.
Angle C (opposite to c) is 40 degrees
c^2 = a^2 + b^2  2 a b cos C
c = 1951.3
The other two angles, A and B, can be obtained using the law of sines.
sin C/c = sin A/a = sin B/b 
Nautical/Maths 
drwls,
If the dimensions are nautical miles, or miles, then the triangle covers a large fraction of the curved earth, and different equations of spherical trigonometry must be used. If that is the case, ignore my previous answer. The radius of the earth must be used to solve the triangle. Someone else may be able to help you.

Nautical/Maths 
Mike,
Thank you drwls,
The accepted format for determining nautical miles is a capital M. I put NM just in case someone was confused. Sorry for the additional confusion.
Now to the problem.
Again the PZX triangle is to do with celestial navigation and I understand spherical trigonometery is appropriate.
I am able to find all the formulas required on line but simply do not understand how to progress.
As with all things I am sure it is simple when you know how.
Thanks
Mike 
Nautical/Maths 
Damon,
One degree of great circle = 60 NM
little letters for sides (expressed as degrees or divided by 60 to give NM), big letters for angles
So side z = 2000/60 = 33.3 deg
and side x =3000/60 = 50 deg
now use law of cosines (spherical version) to find side p
cos p= cos z cos x + sin z sin x cos P
cos p = .836(.643)+.549(.766)(.766)
cos p = .860
p = 30.7 deg
that times 60 = 1841 NM for third great circle distance
then use law of sines to get the other two angles
p = 
Nautical/Maths 
Mike,
Thanks Damon,
This is beginning to make sense.
I have printed your answer for future reference.
Something I am not aware of is what tools do I need to complete these calculations.
Do I need tables and/or a calculator?
Additionally I am unsure how you have arrived at p = .860
I am sorry but these calculations are still a total mystery to me.
I do not understand what to do when I see .836(.643)
I think it is just that I am unaware of procedures.
Should I multiply .836 by .643 ?
Then multiply .549 by .766 by .766 ?
Then add the 2 answers together ?
Thanks
Mike 
Nautical/Maths 
Damon,
Should I multiply .836 by .643 ?
Then multiply .549 by .766 by .766 ?
Then add the 2 answers together ?

Yes, yes, you have it.
I used a TI83 calculator that has sin and cos functions 
Nautical/Maths 
Damon,
By the way, in the old days like when I went sailboat racing, we had to use logs of trig functions to do these products by adding and subtracting logs because we did not have calculators. This was a total mess and resulted in complicated things like using half the angles so that the logs came out right. Give prayers of thanks for calculators.

Nautical/Maths 
Mike,
Thanks again Damon,
Thats very clever and I do understand it.
Please remind me of the law of sines (spherical version) I presume appropriate in these circumstances.
Additionally I have read some people prefer to use haversines for these calculations. I am similarly confused about this and wonder if you can explain the differences.
Thanks
Mike 
Nautical/Maths 
Damon,
The haversines are because of that log problem I mentioned. Forget it. We do not have to multiply and divide by adding and subtracting logs any more. Use a calculator.
same old law of sines
sin A/sin a = sin B/sin b = sin C/sin c 
Nautical/Maths 
Damon,
I suppose I should explain more about the haversines (half angle sines)
The log of a negative number is undefined.
Remember the definition of base ten log:
10^log x = x
Now ten to what power is a negative number?
No such thing/
BUT, here we may very well have sines and cosines of angles greater than 90 degrees. We need to multiply them by adding their logs if we do not have a calculator. However outside of the first 90 degrees, we may have negative sines and cosines. Therefore we use half the angles and the half angle formulas to keep our trig functions between zero and one instead of between 1 and +1
Since calculators can handle trig functions of angles greater than 90, we do not need haversines any more.