Let the incircle of triangle ABC have radius 2 and let it be tangent to BC at D. Suppose BD = 3 and DC = 4. What is the length of the longest side of ABC?
BE=3 and CF=4. And similarly, AE = AF.
Were you aware that the 2 tangent lines from a point outside circle to the tangent points are equal?
You would use Law of Sines to calculate angles ABC and ACB, leaving angle BAC = 180 - ABC - ACB.
Let M = inner circle center. Now work with the 3 inner triangles AMB, AMC and BMC.
Were you aware of this formula:
radius-of-inner-circle = 2[(area-of-triangle) / (perimeter of triangle)] ?
If we let x = AE = AF, then perimeter becomes 2x + 14 and area (using Heron's formula)
becomes sqrt[12(x^2 + 7x)].
This would be another way of solving, NOT needing the Law of Sines.
Hope this helps...