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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Question : In a $\triangle \mathrm{ABC}$, the bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$ meet at $\mathrm{O}$. If $\angle \mathrm{BOC}=142^{\circ}$, then the measure of $\angle \mathrm{A}$ is:
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