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9th Class Mathematics Triangles Question Bank Triangle

  • question_answer
    In the given figure below, ABC is ail isosceles triangle with. The side BA Is produced to D such that \[\mathbf{AB}=\mathbf{AD}\]. If \[\angle ABC={{35}^{{}^\circ }}\], then \[\angle \mathbf{BCD}\] is equal to

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

    B)  \[{{90}^{{}^\circ }}\]          

    C)  \[{{30}^{{}^\circ }}\]                                   

    D)  \[{{60}^{{}^\circ }}\]

    Correct Answer: B

    Solution :

    (b): \[AB=AC=AD\] \[\Rightarrow \]\[\angle ABC=\angle ACB={{35}^{{}^\circ }}\] \[\Rightarrow \]\[\angle BAC={{180}^{{}^\circ }}-{{70}^{{}^\circ }}={{110}^{{}^\circ }}\] \[\Rightarrow \]\[\angle ADC+\angle ACD={{110}^{{}^\circ }}\] \[\therefore \]\[\angle ACD=\frac{{{110}^{{}^\circ }}}{2}={{55}^{{}^\circ }}\] \[\therefore \] \[\angle BCD=\angle ACB+\angle ACD\] \[={{35}^{{}^\circ }}+{{55}^{{}^\circ }}={{90}^{{}^\circ }}\] 


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