Using vectro mehod, prove that in a `/_\ABC, a/(sinA),b/(sinB)=c/(sinC)` where a,b,c are the lenths of the sides opposite to the angles A,B and C respectively of `/_\ABC.

Explanation

Choose A as the origin , the X -axis along the line AB and the Y -axis perpendicular to the X axis through the origin A. Since ` angle A lt 180^(@)` , C is above the X -axis Draw CM perpendicular to X -axis metting it at M.
Even though angle A is drawn as an acute angle , the proof is same even if the angle is obtuse.]
`because l(AB)=c" " therefore B-= (c,0)`
`therefore l(AC)=b " "C-=(bcosA,b sin A)`
`therefore l(CM)=b sin A`
Now select AB along the X-axis is such that B as origin and A is on the negative side of X -axis.
`therefore` B is (0,0) and CB makes an angle of `(pi-B)` with the positive side of X -axis.
`therefore C-=(a cos (pi-B),a sin(pi-B)`
`-= (-a cos , B , a sin B)`
`therefore l (CM) = a sin B`
From (1) and (2) , we get ,
` b sin A = a sin B`
`therefore (a)/(sin A) = (b)/(sin B)`
Similarly , we can prove that `(b)/(sin B)=(c)/(sinC)`
`therefore (a)/(sinA)=(b)/(sin B) =(c)/(sinC)`.