Q1 of 47 Page 1

In Fig. 1, PQ is a tangent at a point C to a circle with center O. if AB is a diameter and ∠CAB = 30°. Find ∠PCA.

Given: PQ is a tangent at a point C with center O and AB is a diameter and ∠CAB = 30°


To find: ∠PCA


Construction: Join OC


Consider △AOB



∠CAO = ∠CAB = 30° .......... [Eqn 1]


Also,


AO = OC [radii of same circle]


∠CAO = ∠OCA [angles opposite to equal sides are equal]


∠OCA = 30° [from eqn 1] ...........[eqn 2]


Now, OC ⏊ PQ


[As The tangent drawn at any point of a circle is perpendicular to the radius through the point of contact]


∠OCP = 90°


∠OCA + ∠PCA = 90°


30° + ∠PCA = 90° [from eqn 2]


∠PCA = 60°

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