Find area of triangle

We note that: 1 + $\frac{r}{R}$ = cos(A) + cos(B) + cos(C) or 1 + $\frac{5}{16}$ =3cos(B). In other words, cos(B) = $\frac{7}{16}$ , which in turn implies that sin(B) = $\frac{3√23}{16}$ . Now, a=b cos(C) + c cos(B) and c=a cos(B) + b cos(A) whence (a+c)=b(cos(A)+cos(C))+(a+c) cos(B)=2bcos(B+(a+c) cos(B)=(a+2b+c)cos(B)=(a+2b+c) $\frac{7}{16}$ Hence, a+c = $\frac{14b}{9}$ and further,s = $\frac{a+b+c}{2}$ = $\frac{23b}{18}$ . Now, 16 = $\frac{abc}{4rs}$ . Substitute, the values of r & s in this formula to obtain, ac= $\frac{3680}{9}$ . We can now claim that: A = ac $\frac{sin(B)}{2}$ = $\frac{115√23}{3}$ using the value of sin(B) obtained earlier.