Question

Show that for any triangle (a^2+b^2+c^2)/(2abc)=(cos A/a)+(cos B/b)+(cos C/c)

Answer 1

expand each square term on left using cosine rule.
(a^2+b^2+c^2)/(2abc) = (b^2+c^2 - 2bc cosA + a^2+c^2 - 2 ac cos B +a^2+b^2 - 2 ab cos C)/2
= 2(a^2 + b^2 + c^2)/2abc -1/2( (cos A/a)+(cos B/b)+(cos C/c) )

let (a^2+b^2+c^2)/(2abc) = x
then
1/2 x = x - 1/2((cos A/a)+(cos B/b)+(cos C/c))
1/2 x = 1/2((cos A/a)+(cos B/b)+(cos C/c))

Answer 2

Use the law of cosines
\[a^2=b^2+c^2-2bc \cos(A)\]
\[b^2=a^2+c^2-2ac \cos(B)\]

\[c^2=a^2+b^2-2ab \cos(C)\]

Answer 3

Combine the fractions on the right hand side of the equal sign

Answer 4

So you will have
\[\frac{\cos(A)}{a}+\frac{\cos(B)}{b}+\frac{\cos(C)}{c}=\frac{bc \cos(A)+ac \cos(B)+ab \cos(C)}{abc}\]
and by the way i know who you are @zotactic :) PS

Answer 5

So you will also need to multiply that fraction by 2/2 since the other side has a 2 in the bottom

Answer 6

\[\frac{2}{2} \cdot \frac{bc \cos(A)+ac \cos(B)+ab \cos(C)}{abc}\]

Answer 7

Distribute the 2 on top

Answer 8

Now go back to the equations I wrote for you at the beginning
and solve each of them for the 2*side*side cos(angle) thingy ma jigger

Answer 9

And replace in the new expression I wrote