Question

In the figure given below, $AB$ and $CD$ are straight lines through the centre $O$ of a circle. If $\angle AOC={80}^o$ and $\angle CDE={40}^o$, find :
i. $\angle DCE$
ii. $\angle ABC$

Answer 1

We show here,

(i) $\angle CED={90}^0$

(Angle in a semicircle is a right angle)

$\begin{array}{c}\therefore \angle DCE={90}^0-\angle CDE={90}^0-{40}^0={50}^0\\ \therefore \angle DCE=\angle OCB={50}^0\end{array}$

(ii) In $\Delta BOC$,

$\angle AOC=\angle OCB+\angle OBC$

(Exterior angle of a $\Delta$ is equal to the sum of pair of interior opposite angle)

$\begin{array}{c}\angle OCB={80}^0-{50}^0={30}^0\left[\angle AOC={80}^{0}given\right]\\ Hence,\angle ABC={30}^0\end{array}$