In the given figure ABCD is a cyclic quadrilateral.The tangent to the circle at B meets DC produced at F. If $\angle EAB=85°$ and $\angle BFC=50°$ , find $\angle CAB$.
+128566 Angle DAC = 180 - 85 = 95°
And since ABCD is a cyclic quadtilateral, then angle DCB = 180 -95 = 85°
Then angle BCF =180 - 85 = 95°
So angle FBC =180 - 50 -95 = 35°
And this = (1/2) measure of minor arc BC = 2 (35) = 70°
But angle CAB is an inscribed angle itercepting this arc....so its measure = (1/2) (70) = 35°
