Question : In a circle with centre O, AB is a diameter and CD is a chord which is equal to the radius OC. AC and BD are extended in such a way that they intersect each other at a point P, exterior to the circle. The measure of $\angle$APB is:
Option 1: 30°
Option 2: 45°
Option 3: 60°
Option 4: 90°
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Correct Answer: 60°
Solution :
In the triangle, OCD is an equilateral triangle as OC = OD = CD (which are all radii of the circle).
Therefore, $\angle$COD = 60°.
$\angle$CBD = $\frac{1}{2}\angle$COD = 30°
Since AB is a diameter, $\angle$ACB = 90° (angle in a semi-circle)
In triangle BCP,
$\angle$BCP = 180° – $\angle$ACB = 180° – 90° = 90° (straight line)
In $\triangle$BCP,
$\angle$APB = 180° – $\angle$BCP – $\angle$CBD = 180° – 90° – 30° = 60°.
Hence, the correct answer is 60°.
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