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 & cosines calculated are positive and I agree with the answer in the book no problem.
In the second example one of the angles is 100 degrees and gives a negative cosine (cosine C). In this case I do not agree with the solution in the book but believe it is that I am unaware what to do with the negative cosine.
Not sure if this will help but triangle CBA
Angle C is known 100 degrees
Length of a is known 34 degrees
Length of b is known 38 degrees
As the sphere is the earth 1 degree = 60 nautical mile.
The answer given for length c is 54 degrees which multiplied by 60 = 3240 nautical miles.
I am unable to discover where 54 degrees comes from but feel it is the negative cosine confusing me.
Please help
Thanks
Mike
Navigator reporting. Dory is bailed out and oars shipped.
I agree with the 54 degrees.
Wicked ice on the harbor this morning. Had to break crust to get to breakfast up at Cripple Cove landing.
Sorry about being late for watch :)
Anyway:
cos c = cos a cos b + sin b sin a cos C
cos c = .829*.788 + .61566*.559 cos C
cos c = .653 + .344 ( -.1736)
cos c = .653 - .0597
cos c = .593
so
c = 53.6 degrees
You probably were messed up by the angle(distance) c being wanted rather than a as the law is usually written.
Best to remember it using this picture in your head:
cos side unknown = cos other side*cos third side + sin other side*sin third side* cos angle opposite unknown side
Thanks again Damon.
Problem is I cannot get my calculator to come up with -.0597 by using .344(-.1736)?
All my figures for sines and cosines are the same as yours but I obviously do not know how to use my calculator!
.344 x -.1736 gives me .1704 on my calculator?
However:!!!!!
-.1736 x .344 = .0597 eurika!
Is it always necessary to enter the negative values first? or is it my calculator?
Thanks again
Mike
Your calculator probably does not like getting two orders at once, like negative and multiply. It was the x- that confused it.
To find the length of side c in your spherical triangle, you can use the law of cosines for spherical triangles. The formula you mentioned,
cos(c) = cos(a) * cos(b) + sin(a) * sin(b) * cos(C),
is the correct formula to use. However, when working with spherical triangles, the angles are measured in degrees and the sides are measured in degrees of arc.
In the example you provided, you have angle C = 100 degrees, and the lengths of sides a and b are given as 34 degrees and 38 degrees respectively. To find the length of side c, you can rearrange the formula above to solve for cos(c) as follows:
cos(c) = (cos(a) * cos(b)) + (sin(a) * sin(b) * cos(C))
Now, plug in the given values:
cos(c) = (cos(34) * cos(38)) + (sin(34) * sin(38) * cos(100))
You may be confused about the negative cosine value because the cosine function can be negative for certain angles. In this case, it means that the angle C is between 90 and 180 degrees. However, the squared lengths of the sides in the spherical law of cosines equation will cancel out any negative sign, so there is no need to worry about the negative cosine value.
Next, you can evaluate the equation using a calculator or a programming language that supports trigonometric functions.
cos(c) = (cos(34) * cos(38)) + (sin(34) * sin(38) * cos(100))
cos(c) = (0.8290375726 * 0.7833269096) + (0.5591929035 * 0.6216099683 * -0.1736481777)
cos(c) = (0.6491615568) + (-0.043473507)
cos(c) = 0.6056880498
To find the length of side c, you can take the inverse cosine (also known as arccos) of the value obtained above:
c = arccos(0.6056880498)
Using a calculator or a programming language, you can find that the result is approximately 53.8935 degrees.
Therefore, the length of side c is approximately 53.8935 degrees of arc, which, as you mentioned, can be converted to nautical miles by multiplying by 60:
c = 53.8935 * 60 = 3233.61 nautical miles.
So, the answer given in the book is approximately correct.