The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ${\angle PCA =110^0}$ , find $ \angle CBA$ .Hint : Join C with centre O.

Let O be the centre of the circle, so AB is a diameter and the radius OC is perpendicular to the tangent at P, making angle OCP 90 degrees. Since angle PCA is 110 degrees, angle OCA is 110 minus 90, which equals 20 degrees. In triangle AOC, OA and OC are radii, so the base angles are equal; therefore, angle OAC (same as angle CAB) is 20 degrees. The angles of triangle ABC must sum to 180, so angle CBA is 180 minus 90 minus 20, giving a measure of 70 degrees.

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