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9th Class Mathematics Lines and Angles Question Bank Lines and angles

  • question_answer
    In the given figure,\[\mathbf{AB}\parallel \mathbf{DC},\]\[\angle \mathbf{BAD}=\mathbf{9}{{\mathbf{0}}^{{}^\circ }},\]\[\angle \mathbf{CBD}=\mathbf{3}{{\mathbf{8}}^{{}^\circ }}\]and\[\angle \mathbf{BCE}=\mathbf{7}{{\mathbf{5}}^{{}^\circ }}\]. Then \[\angle \mathbf{ABD}\] = ?

    A)  \[{{32}^{{}^\circ }}\]                                   

    B)  \[{{37}^{{}^\circ }}\]

    C)  \[{{34}^{{}^\circ }}\]                                   

    D)  \[{{35}^{{}^\circ }}\]

    Correct Answer: B

    Solution :

    (b): \[\angle BCD+\angle BCE={{180}^{{}^\circ }}\Rightarrow \angle BCD+{{75}^{{}^\circ }}={{180}^{{}^\circ }}\] \[\Rightarrow \] \[\angle BCD=({{180}^{{}^\circ }}-{{75}^{{}^\circ }})={{105}^{{}^\circ }}\] In, \[\Delta BCD\] we have : \[\angle CBD+\angle BCD+\angle BDC={{180}^{{}^\circ }}\] \[\Rightarrow \]\[{{38}^{{}^\circ }}+{{105}^{{}^\circ }}+\angle BDC={{180}^{{}^\circ }}\] \[\therefore \]\[\angle BDC={{37}^{{}^\circ }}\] Since \[AB\parallel DC\], we have, \[\angle ABD=\angle BDC={{37}^{{}^\circ }}\] \[\therefore \]\[\angle ABD={{37}^{{}^\circ }}\]


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