Question

In the figure at right, AB and CD are straight lines through the centre O of a circle. If $\angle AOC={98}^o$ and $\angle CDE={35}^o$
Find (i) $\angle DCE$
(ii) $\angle ABC$

Answer 1

Given :

$\angle AOC=98°$

$\angle CDE=35°$

In $△DCE$

$\angle CDE=35°$

$\angle CED=90°$ (angle made with diameter base)

$\therefore \angle DCE=180°-125°=55°$

$\angle COB=180°-98°=82°$

Now $\angle OME=360°-(82°+55°+90°)=133°$

$\angle EMB=180°-133°=47°$

Now in $△MEB$

$\angle MBE=180°-(47°+90°)=43°$

$\therefore \angle ABC=43°$