In \(\triangle ABC\), internal angle bisector of \(\angle BAC\) divides \(\triangle ABC\)'s circumcircle into two arcs whose ratio of lengths is \(7:11\). Given that \(\angle ACB = 36°\), find the measure of \(\angle ABC\) in degrees.
Note that minor arc AB =2 (36°) = 72°
And arc ABM is 7/18 of the circumference = (7/18) * 360 =140°
So minor arc BM = 140 -72 = 68° = 2*angle BAM = angle BAC
So angle ABC = 180 - 36 - 68 = 76°
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