In the adjoining figure, $$P$$ is the centre of the first circle, which touches the other circle in $$C$$. $$PCD$$ is along the diameter of the second circle. $$\angle PBA = 20^\circ$$ and $$\angle PCA = 30^\circ$$.
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The tangents at $$B$$ and $$D$$ meet at $$E$$. The measure of the angle $$x$$ is

PA, PB, PC are the radius of the circle. 

This implies $$\triangle PAB, \triangle PAC$$ are isosceles triangles.

$$\angle BAP = \angle PBA = 20\degree$$

$$\angle PAC =\angle PCA = 30\degree$$

$$\implies \angle BAC = \angle BAP+\angle PAC = 30\degree+20\degree = 50\degree$$

$$\angle BPC = 2\angle BAC = 100\degree$$ ( Angle subtended by a chord at the centre of the circle is twice the angle subtended at the point on the circumference)

Consider quadrilateral BPDE,

$$\angle PBE = \angle PDE = 90\degree$$ (Angle made by the tangent with the radius)

$$ x + \angle PBE + \angle PDE +\angle BPC =  360\degree$$

$$\implies x = 360\degree - (100\degree+90\degree+90\degree) = 80\degree$$ (Angle sum property of a quadrilateral)