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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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:
Question : ABCD is a cyclic quadrilateral such that AB is the diameter of the circle circumscribing it and $\angle $ADC = 118°. What is the measure of $\angle$BAC?
Question : ABCD is a cyclic quadrilateral such that AB is the diameter of the circle circumscribing it and $\angle A D C=148^{\circ}$. What is the measure of the $\angle BAC$?
Question : PA and PB are two tangents from a point P outside the circle with centre O. If A and B are points on the circle such that $\angle \mathrm{APB}=128^{\circ}$, then $\angle \mathrm{OAB}$ is equal to:
Question : PA and PB are two tangents from a point P outside the circle with centre O. If A and B are points on the circle such that $\angle \mathrm{APB}=142^{\circ}$, then $\angle \mathrm{OAB}$ is equal to:
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