Question

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

Answer 1

(i)
$\angle CED={90}^{\circ }$ (Angle in a semi circle)
In $\Delta CED$, we have:
$\angle CED+\angle EDC+\angle DCE={180}^{\circ }$ (Angle sum property of a triangle)
$\Rightarrow {90}^{\circ }+{40}^{\circ }+\angle DCE={180}^{\circ }$
$\Rightarrow \angle DCE=(180–130)={50}^{\circ }$
$\therefore \angle DCE={50}^{\circ }$.... ...(i)
(ii)
As $\angle AOC$ and $\angle BOC$ are linear pair,
we have: $\angle BOC=({180}^{\circ }–{80}^{\circ })={100}^{\circ }$........(ii)
In $\Delta BOC$, we have:
$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$ (Angle sum property of a triangle)
$\angle ABC+\angle DCE+\angle BOC={180}^{\circ }$ [Since, $\angle OBC=\angle ABC$ and $\angle OCB=\angle DCE$]
$\angle ABC={180}^{\circ }–(\angle BOC+\angle DCE)$
$\angle ABC={180}^{\circ }–({100}^{\circ }+{50}^{\circ })$ [From (i) and (ii)]
⇒ $\therefore \angle ABC=({180}^{\circ }-{150}^{\circ })={30}^{\circ }$