Question : In a circle, O is the centre of the circle. Chords AB and CD intersect at P. If $\angle AOD=32^{\circ}$ and $\angle CO B=26^{\circ}$, then the measure of $\angle APD$ lies between:
Option 1: 18º and 22º
Option 2: 26º and 30º
Option 3: 30º and 34º
Option 4: 22º and 26º
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Correct Answer: 26º and 30º
Solution :
The angle made by an arc at the centre of the circle is twice the angle formed at the other part by the same arc.
According to the question
$\angle$ AOD = 32º ; $\angle$ COB = 26º
$\angle$ ABD = $\frac{1}{2}$ of 32º = 16º
$\angle$ CDB = $\frac{1}{2}$ of 26º = 13º
In ${\triangle}$BPD
⇒ $\angle$ DPB = 180º – (13º + 16º) = 151º
Now,
⇒ $\angle$ DPB + $\angle$ APD = 180º
⇒ $\angle$ APD = 180º – 151º = 29º
Hence, the correct answer is 26º and 30º.
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