In the figure, $O$ is the centre of the circle and the length of arc $AB$ is twice the length of arc $BC.$ If angle $AOB = 108^\circ$ find $: \angle ADB.$
Exercise 17 (B) | Q 8.2 | Page 265
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Again, Arc $AB$ subtends $\angle AOB$ at the centre and
$\angle ACB$ at the remaining part of the circle.
$\angle ACB =\frac{1}{2} \angle AOB$
$=\frac{1}{2} \times 108^{\circ}$
$=54$
In cyclic quadrilateral $ADBC$
$\angle ADB + \angle ACB =180^\circ$ (sum of opposite angles)
$\Rightarrow \angle ADB + 54^\circ = 180^\circ$
$\Rightarrow \angle ADB = 180^\circ - 54^\circ$
$\Rightarrow \angle ADB = 126^\circ$
art

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