10th Class Mathematics Solved Paper - Mathematics-2016 Outside Delhi Set-I

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


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