1.

In `DeltaABC`, prove that: `(b+c)cos(B+C)/(2)=acos(B-C)/(2)`

Answer» LHS `=(b+c)cos(B+C)/(2)`
`=(ksinB+ksinC).cos(pi-A)/(2)`
`k.(sinB+sinC).sinA/2`
`k.2sin(B+C)/(2).cos(B-C)/(2).sinA/2`
`k.2sin(pi-A)/(2).sinA/2.cos(B-C)/(2)`
`=k.(2cosA/2sinA/2)cos(B-C)/(2)`
`=k.sinAcos(B-C)/(2)`
`=acos(B-C)/(2)` = RHS Hence Proved.


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