1.

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

Answer» `acos(B-C)+bcos(C-A)+c cos(A-B)`
`=2RsinAcos(B-C)+2RsinBcos(C-A)+2RsinC cos(A-B)`
`=2Rsin(pi-(B+C))cos(B-C)+2Rsin(pi-(C+A))cos(C-A)+2Rsin(pi-(A+B)) cos(A-B)`
`=2Rsin(B+C)cos(B-C)+2Rsin(C+A)cos(C-A)+2Rsin(A+B) cos(A-B)`
`=R(sin2B+sin2C+sin2C+sin2A+sin2A+sin2B)`
`=2R(sin2A+sin2B+sin2C)`
`=2R(4sinAsinBsinC)`
`= 8RsinAsinBsinC`
`= 8R(a/(2R))(b/(2R))(c/(2R))`
`=(abc)/R^2`
`:. acos(B-C)+bcos(C-A)+c cos(A-B) = (abc)/R^2`


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