posted by .

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.


  • Nautical/Maths -

    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 -

    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 -

    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.



  • Nautical/Maths -

    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 -

    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 ?



  • Nautical/Maths -

    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 TI-83 calculator that has sin and cos functions

  • Nautical/Maths -

    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 -

    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.



  • Nautical/Maths -

    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 -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Nautical Studies

    I am just beginning to study celestial navigation and initially wish to concentrate my limited powers on the intercept method. I understand this involves the creation of a spherical triangle PZX which must be solved. I also understand …
  2. Navigation/maths

    I am curious to know what "sight reduction tables" are: I assume they give distance and bearing (azimuth) from one known (or assumed)location on the surface of the earth to another to the accuracy of the tables. Please can someone …
  3. Maths/Law of sines

    I have a spherical triangle and I know 1 angle 31.3 degrees and all 3 sides which are 1624, 2118.4 and 1078.85 nautical miles. In order to find the other 2 angles I know I must use the law of sines: sin A over a = sin B over b = sin …
  4. Spherical Trigonometery

    I am trying to apply the formula cos c = cos a x cos b + sin a x sin b x cos C to find the length of c in my spherical triangle. I am working with 2 examples in a book in which the answers are given. In the first example all the sines …
  5. Navigation

    Just to see if we are talking the same language: Navigation triangle PZX LHA = 194.65 degrees (P) Lat of Z 51.74 degrees Lat of X 39.24 degrees Solve for distance p Thanks Mike
  6. Navigation

    Solution to yesterdays problem I think! I am just beginning to study navigation & on top of that do not have a clue about how things should be expressed mathematically and on top of that have difficulty typing formulas on the computer …
  7. Navigation

    Still studying navigation and am trying to consolidate my knowledge of the navigational triangle. Current interest in particular is how to establish the distance of the unknown side of the PZX triangle. From the assumed position of …
  8. Physics

    Can someone help me get started on this problem?
  9. math

    A ship sails due north from a position 5 degrees, 28' South Latitude to position 6 degrees, 43' North Latitude. Given that one minute of latitude is equivalent to 1 nautical mile, the ship has sailed a distance of A. 75 nautical miles …
  10. Maths

    Please could someone help or complete an unusual calculation for me. I need to solve a triangle where all the angles are known and 2 of the sides. 2 of the sides are identical and 2 of the angles are identical. One angle is one sixtieth …

More Similar Questions