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9th Class Mathematics Circles Question Bank Circle

  • question_answer
    AC is the diameter of the circumcircle of the cyclic quadrilateral ABCD. If. \[\angle \mathbf{BDC}=\mathbf{4}{{\mathbf{8}}^{{}^\circ }}\]then \[\angle \mathbf{ACB}\] is equal to

    A)  \[{{42}^{{}^\circ }}\]      

    B)  \[{{45}^{{}^\circ }}\]           

    C)  \[{{48}^{{}^\circ }}\]                                   

    D)  \[{{58}^{{}^\circ }}\]

    Correct Answer: A

    Solution :

    (a) \[\therefore \]\[\angle ADC={{90}^{{}^\circ }}\] (Angle in semi-circle is a right angle) \[\angle ADB=\angle ADC-\angle BDC={{90}^{{}^\circ }}-{{48}^{{}^\circ }}={{42}^{{}^\circ }}\] \[\angle ADB=\angle ACB={{42}^{{}^\circ }}\] (Angle in a same segment in a right angle)                        


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