Question

In the figure, if $\angle AOB={80}^{o}and\angle ABC={30}^o$, then find $\angle CAO$.

Answer 1

In the figure, $\angle AOB={80}^o,\angle ABC={30}^o$

$\because$ Arc AB subtends $\angle AOB$ at the centre and $\angle ACB$ at the remaining part of the circle.

$\therefore \angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {80}^o={40}^o$

In $\Delta OAB,OA=OB$

$\therefore \angle OAB=\angle OBA$

But, $\angle OAB+\angle OBA+\angle AOB={180}^o$

$\therefore \angle OAB+\angle OBA+{80}^o={180}^o$

$\Rightarrow \angle OAB+\angle OAB={180}^o-{80}^o={100}^o$

$\therefore 2\angle OAB={100}^o$

$\Rightarrow \angle OAB=\frac{{100}^o}{2}={50}^o$

Similarly, in $\Delta ABC$,

$\angle BAC+\angle ACB+\angle ABC={180}^o$

$\angle BAC+{40}^o+{30}^o={180}^o$

$\Rightarrow \angle BAC={180}^o-{30}^o-{40}^o$

$={180}^o-{70}^o={110}^o$

$\therefore \angle CAO=\angle BAC-\angle OAB$

$={110}^o-{50}^o={60}^o$