Posted by **mathstudent** on Friday, December 15, 2006 at 4:19pm.

There is an arbitrary triangle with angles A, B, and C and sides of lengths a, b, and c. Angle A is opposite side a.

How do I get the formulas:

b * cos C + c * cos B = a

c * cos A + a * cos C = b

a * cos B + b * cos A = c

Are these standard trig formulas? What are they called? Or where are they derived from?

thanks!

b * cos C + c * cos B = a

Draw the line at right angles to side a that splits the angle A (not necessarily in equal parts). Then, the triangle is slit in two right triangles, and you see that

a = a1 + a2

And:

a1 = b * cos C

a2 = c * cos B

Cyclically permuting a -- > b---> c ---> a (and the uppercase variables) give you the other two equations)

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