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 ∠APB is:
Show answer & explanation
Correct answer: C
Explanation
The correct answer is Option 3 i.e. 60
| Application |
|
Here we draw a diagram:
|
|
Now, Since CD = OC = radius So, ∆ODC is an equilateral triangle ∠OCD = ∠ODC = ∠COD = 60º We know In a circle, the angle at the center which is formed by the same arc is twice the angle at the circumference. So, ∠CAD = 60º/2 = 30º |
|
According to circle rule: AD is a perpendicular to BD. So, ∠ADB = 90º = ∠ADP |
|
Hence, From the triangle rule: Sum of all angle = 180º So, ∠PAD + ∠ADP + ∠APD = 180º ⇒ 30º + 90º + ∠APD = 180º ⇒ ∠APD = 180º - 120º ⇒ ∠APD = 60º ⇒ ∠APB = 60º |
