Ans is B 30

since AB is equal to the radius of the circle triangle OAB is equilateral. hence angle AOB is 60
x=1/2*angle AOB=30

liferocks wrote:Ans is B 30

since AB is equal to the radius of the circle triangle OAB is equilateral. hence angle AOB is 60
x=1/2*angle AOB=30

I understand the equilateral part. But, how is x=1/2*angle AOB=30. Which property is this? Yes, the answer is B

Here, the AB is equal to the Radius. So if we draw a triangle with AB as the base and the third vertex as the centre O, we get a triangle AOB. Now Angle x and Angle AOB is made by the same base AB.

Now we know angle at the centre of the circle is twice the angle at the circumference.
or, angle AOB = 2* angle x
Now, triangle AOB is equilateral.

60 = 2*x
or, x=30

Thus B

Nilakshi1001 wrote:Here, the AB is equal to the Radius. So if we draw a triangle with AB as the base and the third vertex as the centre O, we get a triangle AOB. Now Angle x and Angle AOB is made by the same base AB.

Now we know angle at the centre of the circle is twice the angle at the circumference.
or, angle AOB = 2* angle x
Now, triangle AOB is equilateral.

60 = 2*x
or, x=30

Thus B

I still don't get the part about the angle at the center of the circle is twice the angle at the circumference. .I am doing math after 8 years..I want to know the theory ..I know that inscribed angle of a diameter is 90 degrees.

this is the theory behind 'the angle at the center of the circle is twice the angle at the circumference'.Let me know if it clarifies the questions.
here angle AOB=2(x+y)=angle ACB

Simply superb : https://gmatclub.com/forum/math-circles-87957.html