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) \[\therefore \] \[\angle PCA=\angle PCO-\angle ACO\] \[=90{}^\circ -30{}^\circ =60{}^\circ \]
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