Let , #vec(AB)=bar(c) ,# #vec(BC)=bar(a) ,# #vec(CA)=bar(b) #
. #
So,
#bar(a)+bar(b)+bar(c)=bar(0)#
Using definition of cross Product
#bar(a)xx(bar(a)+bar(b)+bar(c))=bar(a)xxbar(0)#
#=>(bar(a)xxbar(a))+(bar(a) xxbar(b))+(bar(a) xxbar(c))=bar(0)to[becausebar(a)xxbar(0)=bar(0) ]#
#=>bar(0)+(bar(a) xxbar(b))+(bar(a) xxbar(c))=bar(0)to[because(bar(a)xxbar(a))=bar(0)]#
#=>(bar(a) xxbar(b))-(bar(c)xxbar(a))=bar(0)#
#=>(bar(a) xxbar(b))=(bar(c)xxbar(a))#
#=>a*bsin(pi-C)=c*asin(pi-B)#
#=>bsinC=csinB#
#=>sinC/c =sinB/b....to(1)#
Similarly we can prove that ,
#=>sinA/a =sinB/b....to(2)#
Hence ,
#sinA/a =sinB/b=sinC/c #