In triangle `ABC,` if `A-B=120^@ and R=8r,` where `R and r` have their usual meanings, then `cos C` equal (a) `3/4` (b) `2/3` (c) `5/6` (d) `7/8`

Explanation

`R = 8r`

`=>R = 8(4Rsin(A/2)sin(B/2)sin(C/2))`

`=>2sin(A/2)sin(B/2)sin(C/2) = 1/16`

`=>(cos((A-B)/2) - cos((A+B)/2))sinC/2 = 1/16`

As, `(A-B) = 120^@ and C = pi-(A+B),`

so, our equation becomes,

`=>(1/2-sin(C/2)) sin(C/2) = 1/16`

`=>sin(C/2) = 1/4`

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