This one is a little difficult, Solveit
Look at the following illustration........
On CB, draw BE = BF = 5
And arccos(CBA) = 5/18
And we can bisect this angle and form kite BFDE........where triangles BFD and BDE are congruent right triangles by SAS.......note that DF = DE and DEB and DFB are right angles, so any inscribed circle will have a radius that is tangent to CB and AB at E and F
And (1/2) of angle CBA = (1/2)arccos(5/18)
The x coordinate for the center of the circle = 0.... and the y coordinate lies on the bisector (BD) of base angle CBA at point D and is given by :
tan [ (1/2)arccos(5/18)] = y / 5
5 tan [ (1/2)arccos(5/18)] = y = DF = 3.75904705778
And this is the approximate radius of the inscribed circle......and DF = DE = 3.75904705778
