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Lines and Angles — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSLines and Angles1 Mark
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Answer

  1. 35º
    Solution:
    $\angle\text{ABC}=\angle\text{BCD}=45^\circ$ (Alternate interior angles)
    In $\triangle\text{COD}$
    $\angle\text{COD}+\angle\text{CDO}+\angle\text{DCO}=180^\circ$ (Angle sum property)
    $\angle\text{CDO}=180^\circ-100^\circ-45^\circ$
    $\angle\text{CDO}=35^\circ.$

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