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 $: \angle CAB$
Exercise 17 (B) | Q 8.1 | Page 265
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Join $AD$ and $DB$
Arc $B = 2$ arc $BC$ and $\angle AOB = 180^\circ$
$\therefore \angle BOC = 1 \angle AOB$
$=\frac{1}{2} \times 108^{\circ}$
$=54^{\circ}$
Now, Arc $BC$ subtends $\angle BOC$ at the centre and
$\angle CAB$ at the remaining part of the circle.
$\therefore \angle C A B=\frac{1}{2} \angle B O C$
$=\frac{1}{2} \times 54^{\circ}$
$=27^{\circ}$
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