Question

In the given figure , O is the center of the circle. If $\angle AOB={140}^o$ and $\angle OAC={50}^o$; find :
$\left(i\right)\angle ACB,$
$\left(ii\right)\angle OBC,$
$\left(iii\right)\angle OAB,$
$\left(iv\right)\angle CBA.$

Answer 1

$\begin{array}{c}\left(1\right)\angle ACB=\frac{1}{2}\angle AOBbi\sec tors\\ =\frac{1}{2}\times \left({360}^0-{140}^0\right)=\frac{1}{2}\times 220={110}^0\\ \left(2\right)\angle OBC={360}^0-\left(\angle AOB+\angle OAC+\angle ACB\right)\\ ={360}^0-\left({140}^0+{50}^o+{110}^0\right)\\ ={360}^0-{300}^0={60}^0\\ \left(3\right)\sin ce,OA\&OBisradius\\ \therefore \Delta OABisisosceles\Delta \\ \therefore \angle OAB=\frac{1}{2}\left({180}^0-{140}^0\right)=\frac{1}{2}\times {40}^0={20}^0\\ \left(4\right)\angle CBA-\angle OBC-\angle OBA\\ =\angle OBC-\angle OAB\left[\Delta OABisisosceles\Delta \right]\\ ={60}^0-{20}^0-{40}^0\end{array}$