Question 84

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

Solution

$$\triangle OCD is an equilateral triangle ,

$$\angle COD = 60 \degree$$

$$\angle CBD = \frac{1}{2} \angle COD = 30 \degree$$

$$\angle ACB = 90\degree (\angle ACB is an angle of semi circle. )$$

$$\angle PCB = 90\degree$$

$$\angle PBC = 180\degree - 90\degree - 30\degree = 60\degree$$

So, the answer would be option c)$$60^\circ$$

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