The tangent at pt C of a circle and a diameter AB when extended intersect at P.If angle PCA = 110 find angle CBA



 

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Let 'D' be the center of the circle. So, points A,D,B,P all are on the same line. Also, P and C are points on the tangent.
Join CD and BC.
Now, ∠BCA is the angle inscribed in a semi-circle, so it's 90º.
C is the point on the circle where the tangent touches the circle. So, ∠DCP = 90º.
∠PCA =∠PCD + ∠DCA
110º = 90º + ∠DCA
So, ∠DCA =20º
In triangle ADC, AD = DC (Both are radii of the circle)
So, ∠DCA = ∠CAD =20º

In triangle ABC: ∠BCA = 90º, ∠CAB = 20º
So, ∠CBA = 70º.

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