Question

In the given figure, O is the centre of a circle and chords AC and BD intersect at E. If $\angle AEB={110}^{\circ }and\angle CBE={30}^{\circ }then\angle ADB=?$

(a) ${70}^{\circ }$

(b) ${60}^{\circ }$

(c) ${80}^{\circ }$

(d) ${90}^{\circ }$

Answer 1

ANSWER:
( c ) 80°
We have:
$\angle AEB+\angle CEB$ = 180 ° (Linear pair angles)
⇒ 110 ° + $\angle CEB$ = 180 °
⇒ $\angle CEB$ = (180° - 110°) = 70°
In Δ CEB, we have:

$\angle CEB+\angle EBC+\angle ECB$ = 180° (Angle sum property of a triangle)
⇒ 70° + 30° + $\angle ECB$ = 180°
⇒ $\angle ECB$ = (180° - 100°) = 80°
The angles in the same segment are equal.
Thus, $\angle ADB$ = $\angle ECB$ = 80°