Question : AB is a chord in a circle with centre O. AB is produced to C such that BC is equal to the radius of the circle. C is joined to O and produced to meet the circle at D. If $\angle \mathrm{ACD}=32^{\circ}$, then the measure of $\angle \mathrm{AOD}$ is _____.
Option 1: 48°
Option 2: 96°
Option 3: 108°
Option 4: 80°
Correct Answer: 96°
Solution :
In $\triangle$OBC
OB = BC
So, $\angle$ BOC = $\angle$ BCO = 32°
Also, $\angle$ OBA = $\angle$ BOC + $\angle$ BCO = 32° + 32° = 64°
Since OA = OB
$\angle$ OAB = $\angle$ OBA
In $\triangle$ AOB
$\angle$ AOB + $\angle$ OAB + $\angle$ OBA = 180º
⇒ $\angle$ AOB + 64° + 64° = 180°
⇒ $\angle$ AOB = 180° – 128° = 52°
Now,
$\angle$ AOD + $\angle$ AOB + $\angle$ BOC = 180°
⇒ $\angle$ AOD + 52° + 32° = 180°
⇒ $\angle$ AOD = 180° – 84° = 96°
Hence, the correct answer is 96°.
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