With the help of sinus law that establishes
a/sin(hat A) = b/sin(hat B) = c/sin(hat C)
and also
hatA + hatB + hatC = pi
The formulation is
{
(a/sin(hat A) = b/sin(hat B)),
(b/sin(hat B) = c/sin(hat C)),
(hatA + hatB + hatC = pi)
:}
By knowing that
sin(pi-(hat A+ hat B))=sin(hat A+ hat B)
This system can be reduced to
{
(a/sin(hat A) = b/sin(hat B)),
(b/sin(hat B) = c/sin(hat A + hat B))
:}
Solving for hat A, hat B after substituting
sin(hat A+ hat B) = sin(hat A)cos(hat B)+sin(hat B)cos(hat A)
and cos(alpha) = sqrt(1-sin(alpha)^2) we have:
hat A = arctan((b^2+c^2-a^2)/(bc), sqrt((b+c-a)(a+b-c)(a-b+c)(a+b+c))/(bc))
hat B = arctan((a^2-b^2+c^2)/(ac),sqrt((b+c-a)(a+b-c)(a-b+c)(a+b+c))/(ac)
or substituting values
hat A = arctan(sqrt(7)/3) =41.41^@ , hat B = arctan(5 sqrt(7)/9)=55.77^@ and hat C = 180^@-(41.41^@ +55.77^@) = 82.82^@
