Question : Two chords AB and CD of a circle with centre O intersect at P. If $\angle$ APC = 40º, then the value of $\angle$AOC + $\angle$BOD is:
Option 1: 50º
Option 2: 60º
Option 3: 80º
Option 4: 120º
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Correct Answer: 80º
Solution :
The angle subtended by a chord at the centre is twice the angle subtended on the circumference.
$\angle$BOD = 2$\angle$BCD
⇒ $\angle$AOC = 2$\angle$ABC
⇒ $\angle$AOC + $\angle$BOD = 2($\angle$BCD + $\angle$ABC)
The sum of the interior angle is equal to the exterior angle.
$\angle$BCD + $\angle$ABC = $\angle$APC
⇒ $\angle$AOC + $\angle$BOD = 2$\angle$APC
⇒ $\angle$AOC + $\angle$BOD = 2 × 40º
⇒ $\angle$AOC + $\angle$BOD = 80º
Hence, the correct answer is 80º.
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