Question

Prove that:
cos alpha/a +cos beta / b +cos gamma / c = a^2+b^2+c^2/2abc

Answer 1

\[\frac{ \cos \alpha }{ a }+\frac{ \cos \beta }{ b }+\frac{ \cos \gamma }{ c }=\frac{ a^2+b^2+c^2 }{ 2abc }\]

Answer 2

Reference:-
\[\cos \alpha=\frac{ b^2+c^2-a^2 }{ 2bc }\]

Answer 3

\[\cos \beta=\frac{ a^2+c^2-b^2 }{ 2ac }\]

Answer 4

\[\cos \gamma=\frac{ a^2+b^2-c^2 }{ 2ab }\]

Answer 5

Using L.H.S :-
\[\frac{ \cos \alpha }{ a }+\frac{ \cos \beta }{ b }+\frac{ \cos \gamma }{ c }\]

Answer 6

so,
(b^2+c^2-a^2)/(2abc)+(a^2+c^2-b^2)/(2abc)+(a^2+b^2-c^2)/(2abc)
(a^2+b^2+c^2)/(2abc)
proof ^_^

Answer 7

0_0

Answer 8

whoa!! that was easy!!

Answer 9

yes, its very easy :)