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Lines and Angles — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSLines and Angles1 Mark
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Answer

  1. 37º
    Solution:
    In $\triangle\text{DBC}$
    $\angle\text{BCE}=\angle\text{DBC}+\angle\text{BDC}$ (Exterior angle property)
    $65^\circ=28^\circ+\angle\text{BDC}$
    $\text{BDC}=37^\circ$
    As, AB is parallel to CD
    $\angle\text{ABD}=\angle\text{BDC}=37^\circ$ (Alternate interior angle).

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