MCQ
In the given figure, $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ.$ Then, $\angle\text{CAO}=?$
  • A
    30°
  • B
    45°
  • C
    60°
  • D
    90°

Answer

  1. 60°
    Solution:
    $\angle\text{AOB}=2\angle\text{ACB}$
    $\Rightarrow\ \angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$
    $\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(90^\circ)$
    $\Rightarrow\ \angle\text{ACB}=45^\circ$
    $\angle\text{COA}=2\angle\text{CBA}=2(30^\circ)=60^\circ$
    Since AOD is a straight line,
    $\therefore\ \angle\text{COD}+\angle\text{AOC}=180^\circ$
    $\therefore\ \angle\text{COD}+60^\circ=180^\circ$
    $\therefore\ \angle\text{COD}=120^\circ$
    $\Rightarrow\ \angle\text{CAO}=\frac{1}{2}\angle\text{COD}=\frac{1}{2}\times120^\circ=60^\circ$

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