>>11934509Well, who ever came up with
>>11932638 is telling you how they quite literally did. One way to think about it is that cosine() and sine() are the abscissa and ordinate set that we call (x,y); in this case just x(A), x(B), x(C). Lets look at the ((b^2)+(c^2)-(a^2)) over ((2)(b)(c)).
sine theta = y
cosine theta = x
tangent theta = ((y) / (x)),
_________
cos(A) => x(A)
cos(A) = ((b^2)+(c^2)-(a^2)) / ((2)(b)(c))
x(A) =
_________
The basic set up is this, "a" over "b", of course, but what is "a", or even what is "b" in this case? Well, "a" is made out of our handy distance formula.
distance = (c^2) = ((a^2) + (b^2))
By this, I mean, think about it; the correlation that it has, so it must be a manipulation of the distance formula. What changes between the example of cos(A) and the distance formula is that the distance formula is made of one line, where we have three lines that create the triangle, therefore the ((a^2)) becoming negative is exactly like the ((c^)) being moved over to the other side to have one side equal zero.
((b^2)+(c^2)-(a^2)) = 0
((b^2)+(c^2)) = (a^2)
Which is useful when you want to find what line you are starting with, which in this last example was "(a^2)".
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As for the denominator ((2)(b)(c)), we can think of these has the other to lines that come out of a shared vertex to touch the end points of that line.
/\
/ \
b / \ a
/ \
/ \
/ \
/______\
c
*this is just for a visual; always draw.*
The angle AB is also a contact, so that's one.
The angle AC is also a contact do that's two.
There's the unknown length of "b".
And there's the unknown length of "c".
And look at that.
( (2) (a) (b) )
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Then you do it two more times and then put it all together.
