Question

In the given figure, O is the centre of a circle and $\angle AOB={140}^{\circ }.Then\angle ACB=?$

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

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

(c) ${110}^{\circ }$

(d) ${40}^{\circ }$

Answer 1

(c) 110°
Join AB.
Then chord AB subtends $\angle AOB$ at the centre and $\angle ADB$ at a point D of the remaining parts of a circle.
∴$\angle AOB$ = 2$\angle ADB$
⇒$\angle ADB$=1/2$\angle AOB$=1/2×140°=70°
In the cyclic quadrilateral, we have:
$\angle ADB$ + $\angle ACB$ = 180°
⇒ 70° + $\angle ACB$ = 180°
∴$\angle ACB$ = (180° - 70°) = 110°