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The angle α is 30°.
First draw the radii CK and EJ to the tangency points. Because of the two right angles and equal radii, EC must be parallel to JK and hence ∠CEA=α.
Notice that we didn’t use the fact that the triangle is isosceles, so strictly speaking the problem is over-constrained: the top angle would still be 30° in a non-isosceles triangle.