Question

In the given figure, O is the centre of a circle and $\angle AOB={130}^{\circ }.Then,\angle ACB=?$

(a) ${50}^{\circ }$

(b) ${65}^{\circ }$

(c) ${115}^{\circ }$

(d) ${155}^{\circ }$

Answer 1

ANSWER:
( c ) 115°
Join AD and BD.

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$\angle ADB$
⇒$\angle ADB$=$\frac{1}{2}$$\angle AOB$=$\frac{1}{2}\times 130°$=65°
In cyclic quadrilateral, we have:
$\angle ADB$ + $\angle ACB$ = 180°
⇒ 65 ° + $\angle ACB$ = 180°
∴ $\angle ACB$ = (180° - 65°) = 115°