question_answer

In fig. 1, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and \[\angle CAB=30{}^\circ \], find \[\angle PCA\].

Answer:

Given, \[\angle CAB=30{}^\circ \] and PQ is a tangent at a point C to a circle with centre O.

Since, AB is a diameter.

\[\therefore \]                  \[\angle ACB=90{}^\circ \]

Join OC

\[\angle CAO=\angle ACO=30{}^\circ \,\,(~OA=OC)\]

and, \[\angle PCO=90{}^\circ \] (Tangent is perpendicular to the radius through the point of contact)

\[\therefore \]                              \[\angle PCA=\angle PCO-\angle ACO\]

\[=90{}^\circ -30{}^\circ =60{}^\circ \]