Wednesday

April 1, 2015

April 1, 2015

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

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 ...

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...

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 - This is the problem given: The side of a hill faces due south and is ...

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 ...

cosines - I don't know how to reverse cosine decimals into angles. Can you ...

maths - prove that the length of the perpendicular from the origin to the ...