In a `DeltaABC,` `sin A sinB sinC` is

Options

A. `ge(3sqrt3)/(8)`
B. `le(3sqrt3)/(8)`
C. `le(sqrt3)/(8)`
D. `le(3)/(8)`

Answer

B

Explanation

We have,
`sin^(2)A+sin^(2)B+sin^(2)C`
`=(1-cos2A)/(2)+(1-cos2B)/(2)+sin^(2)C`
`=1/2{cos2A+cos2B}+sin^(2)C`
`=1-cos(A+B)cos(A-B)+sin^(2)C`
`=2-cos^(2)C+cosCcos(A-B)`
`le2-(cos^(2)C-cosC)" "[becausecos(A-B)lt1]`
`=2-(cos^(2)C-cosC)`
`=2-(cosC-(1)/(2))^(2)+1/4=9/4-(cosC-(1)/(2))^(2)le9/4`
`thereforesin^(2)A+sin^(2)B+sin^(2)Cle9/4" "...(i)`
`Now, " "A.M.geG.M.`
`implies(sin^(2)A+sin^(2)B+sin^(2)C)/(3)ge(sin^(2)A sin^(2)B sin^(2)C)^(1//3)`
`implies(sin^(2)A sin^(2)Bsin^(2)C)^(1//3)le3/4" "["Using(i)"]`
`impliessinAsinB sinC le((3)/(4))^(3//2)`
`impliessinAsinB sinC le(3sqrt3)/(8)`