In ΔABC,ifsinAcsinB+sinBc+sinCb=cab+bac+abc, then the value of angle A, is (all symbols used have their usual meaning in a triangle):
We have,
sinAcsinB+sinCc+sinCb=cab+bac+abc
We know that
Using SINErule and we get,
sinAa=sinBb=sinCc=k
sinA=ak......(1)
sinB=bk.......(2)
sinC=ck.......(3)
Now,
akckb+kbc+kcb=cab+bac+abc
abc+kbc+kcb=cab+bac+abc
kbc+kcb=cab+bac
k(bc+cb)=1a(cb+bc)
k=1a
Put the value of k in equation (1) and we get,
sinA=a1a
sinA=1
sinA=sin90o
A=90o
Hence, this is the answer.