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Lines and Angles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsLines and Angles1 Mark
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Answer

  1. 80º
    Solution:
    In BDC
    $\angle\text{BDC}+\angle\text{BCD}+\angle\text{DBC}=180^\circ$
    $\text{BD}\perp\text{AC}$
    $\angle\text{BCD}=90^\circ,\angle\text{DBC}=40^\circ$
    $90^\circ+\angle\text{BCD}+40^\circ=180^\circ$
    $\angle\text{BCD}+130^\circ=180^\circ$
    $\angle\text{BCD}=180^\circ-130^\circ$
    $\angle\text{BCD}=50^\circ$
    $\angle\text{AEB}=\angle\text{CAE}+\angle\text{C}$ (exterior angle)
    $\angle\text{CAE}=30^\circ$
    $\angle\text{C}=50^\circ$
    $\angle\text{AEB}=30^\circ+50^\circ$
    $\angle\text{AEB}=80^\circ.$

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