Question : ABCD is a cyclic quadrilateral. The tangents to the circle at the points A and C on it, intersect at P. If $\angle\mathrm{ABC}=98^{\circ}$, then what is the measure of $\angle \mathrm{APC}$ ?
Option 1: 22°
Option 2: 26°
Option 3: 16°
Option 4: 14°
Correct Answer: 16°
Solution :
$\angle$ABC = 98º
$\angle$ABC + $\angle$CDA = 180º (Property of cyclic quadrilateral)
⇒ $\angle$CDA = 180º – 98º = 82º
So, $\angle$AOC = 2 × $\angle$ADC (the angle at the centre is double that on the circumference by the same arc)
= 2 × 82º
= 164º
Also, $\angle$PAO = $\angle$PCO = 90º
So, $\angle$APC = 180º - 164º
= 16º
Hence, the correct answer is 16º.
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