# Thread: Competitive Exam Q related to Circles

1. ## Competitive Exam Q related to Circles

Hey, I can't solve this Q which is in the prep book of a competitive exam that I'm taking. I have attached a pic of it.

Please do provide an explanation on how the answer can be obtained... Thanks in advance.

2. ## Re: Competitive Exam Q related to Circles

Originally Posted by SevenAces
Hey, I can't solve this Q which is in the prep book of a competitive exam that I'm taking. I have attached a pic of it.
Find the measures of $\text{arc}(AC)~\&~\text{arc}(BC)$.
The measure of $\text{arc}(AB)$ is the angle you require.

3. ## Re: Competitive Exam Q related to Circles

Originally Posted by Plato
Find the measures of $\text{arc}(AC)~\&~\text{arc}(BC)$.
The measure of $\text{arc}(AB)$ is the angle you require.
How do I do that???

4. ## Re: Competitive Exam Q related to Circles

Originally Posted by SevenAces
How do I do that???
Look up measures of arcs in your textbook.

5. ## Re: Competitive Exam Q related to Circles

Originally Posted by Plato
Look up measures of arcs in your textbook.
There aren't any measures of arcs specified..!?!? Can you please explain me step-by-step on how to obtain the solution. I have my test tomorrow!

6. ## Re: Competitive Exam Q related to Circles

Hi SevenAces.
Google angles between secants then regentsprep.org

7. ## Re: Competitive Exam Q related to Circles

Originally Posted by SevenAces
Hey, I can't solve this Q which is in the prep book of a competitive exam that I'm taking. I have attached a pic of it.

Please do provide an explanation on how the answer can be obtained... Thanks in advance.

Angle CBA = Angle ACT = 48 degree

Triangle BCT,
Angle ACB = 180 - 48 - 36 - 48 = 48 degree

Angle subtended by AB at the centre of circle = 2 (48) = 96 degree