• # 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 $\angle PCA$.

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