If `A ,Ba n dC` are the vetices of a triangle `A B C ,` then prove sine rule `a/(sinA)=b/(sinB)=c/(sinC)dot`

Explanation

Let `vec(BC)=veca,vec(CA)=vecb,vec(AB),vec c," so that" veca + vecb=-vecc`. Therefore,
`vecaxxveca+vecaxxvecb=-vecaxxvecc`
`vec0+vecaxxvecb=veccxxveca`
`|vecaxxvecb|=|veccxxveca|`
` ab sin(180^(@)-c)=casin (180^(@)-B)`
`ab sin C =sa sin B`
Dividing both sides by abc, we get
`sinC/c=sinB/b`
`b/sinB=c/sinC`
similarly , `c/sinC= a/sinA`
From (i) and (ii) ,we have
`a/sinA=b/sinB=c/sinC`