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Circles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsCircles1 Mark
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Answer

  1. 60º
    Solution:
    We have:
    $\angle\text{AOB}=2\angle\text{ACB}$
    $\Rightarrow\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=(\frac{1}{2}\times90^\circ)=45^\circ$
    $\Rightarrow\angle\text{ACB}=45^\circ$
    $\angle\text{COA}=2\angle\text{CBA}=(2\times30^\circ)=60^\circ$
    $\therefore\angle\text{COD}=180^\circ-\angle\text{COA}=(180^\circ-60^\circ)=120^\circ$
    $\Rightarrow\angle\text{CAO}=\frac{1}{2}\angle\text{COD}=(\frac{1}{2}\times120^\circ)=60^\circ$
    $\Rightarrow\angle\text{CAO}=60^\circ$

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