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Lines and Angles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsLines and Angles1 Mark
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Answer

  1. 30º
    Solution:
    In $\triangle\text{OEB}$
    $\angle\text{OEB}+\angle\text{EBO}+\angle\text{BOE}=180^\circ$ (Angle sum property)
    $75^\circ+55\circ+\angle\text{BOE}=180^\circ$
    $\angle\text{BOE}=50^\circ$
    $\angle\text{BOE}=\angle\text{COD}=50^\circ$ (vertically opposite angle)
    In $\triangle\text{ODC}$
    $\angle\text{ODC}+\angle\text{DOC}+\angle\text{DCO}=180^\circ$
    $\angle\text{ODC}=180^\circ-100^\circ-50^\circ$
    $\angle\text{ODC}=30^\circ$

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