Posted by **ashley** on Tuesday, December 12, 2006 at 5:45pm.

Show that any triangle with standard labeling...

a^2+b^2+c^2/2abc = cos(alpha)/a + cos(beta)/b + cos(gamma)/c

I don't get it. Can someone please help me.

Start here with the law of cosines:

a^2 = b^2 + c^2 -2bc Cos A

b^2 = a^2 + c^2 -2ac Cos B

c^2 = a^2 + b^2 - 2ab Cos C

Add the three equations. divide both sides by 2abc

## Answer This Question

## Related Questions

- Trig. - I have answers for these problems, but I wanted to check if I had them ...
- Maths/Right Triangle - There must be a relationship (formula) between the ...
- Algebra II/ Trig - Should the triangle be solved beginning with Laws of Sines or...
- Math - Trig - What familiar formula can you obtain when you use the third form ...
- Trig - check my answers plz! - 1. (P -15/17, -8/17) is found on the unit circle...
- math - This is the problem given: The side of a hill faces due south and is ...
- trig - I need to prove parallelogram law using the law of cosines. 2AB^2 + 2BC^2...
- math - How would you establish this identity: (1+sec(beta))/(sec(beta))=(sin^2(...
- tigonometry - expres the following as sums and differences of sines or cosines ...
- Calculus - Hello! I need help with this problem, its been irritating me for ...

More Related Questions