With usual notations, in triangle `A B C ,acos(B-C)+bcos(C-A)+c"cos"(A-B)` is equal to `a b c//R^2` (b) `(a b c)/(4R^2)` `(4a b c)/(R^2)` (d) `(a b c)/(2R^2)`

Options

A. `(abc)/(R^(2))`
B. `(abc)/(4R^(2))`
C. `(4 abc)/(R^(2))`
D. `(abc)/(2R^(2))`

Answer

A

Explanation

`a cos (B -C) + b cos (C -A) + c co s(A -B)`
`2R sin A co s(B -C) + 2R sin B cos (C - A) + 2R sin C cos (A -B)`
`=2R sin (B+ C) cos (B -C) + 2R sin (A + C) .cos (C -A) + 2R sin (A +B) .cos(A -B)`
`=R [sin 2B + sin 2C + sin 2C + sin 2A + sin 2A + sin 2B]`
`= 2R (sin 2A + sin 2B + sin 2C)`
`=8R sin A sin B sin C`
`=8R (a)/(2R) (b)/(2R)(c)/(2R) = (abc)/(R^(2))`