# A circle is circumscribed around triangle ABC, find length?

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1. Jun 12, 2017

### Helly123

• New poster has been reminded to use the Homework Help Template when starting threads in the schoolwork forums

some formula related

I tried to draw the problem

can anyone give me clue how to solve it?

2. Jun 12, 2017

### QuantumQuest

Hi Helly123

In order to be helped you must first tell us what is asked. Also, think about it for a while and tell how do you think you must go about solving it (at least in rough lines).

3. Jun 12, 2017

### Buffu

Arc-length is $l = r \theta$.

4. Jun 12, 2017

### Helly123

the question : (1) how the length of arcs AB, BC, CA, expressed in a, b, c, r.
(2) I don't know what kind of expression in number 2
(3) find length of AB, BC, CA, when a = 75 degrees, b = 60, c = 45, r = 1. without trigonometric function

5. Jun 12, 2017

### Helly123

can you give me more clue.....

6. Jun 12, 2017

### Buffu

Can you express arclengths in terms of $a,b,c ,r$ ?

7. Jun 12, 2017

### Helly123

maybe like
AB / 2phi R = c / 360 degrees
AB = c/360 . 2 phi R

8. Jun 12, 2017

### Buffu

No I did not get what you did. I guess finding arclengths confused you, can you find the angle subtended by each side at centre ?

Also in the question what $\bbox[5px,Border: 2px solid black]{2-2}$ mean ? what does that number mean ? I have never seen such a weird way to denote a unknown.

9. Jun 12, 2017

### Helly123

Maybe the way i draw the circle and triangle is wrong?

10. Jun 13, 2017

### Helly123

That's, the first time i saw it too...

" can you find the angle subtended by each side at centre ? " how to find that?

11. Jun 13, 2017

### Buffu

Find theta.

12. Jun 13, 2017

### Helly123

13. Jun 13, 2017

### cnh1995

Looks good.

Next time onwards, use the upload button to directly add the images in the post.

14. Jun 13, 2017

### Helly123

Ok thanks, im using phone so i sent link. Usually i upload. (Cant upload through phone) btw sir, my answer is right? And the formula too?

15. Jun 13, 2017

### cnh1995

Yes.

I'm no sir..just a guy like you, hanging out here!

16. Jun 13, 2017

### Buffu

Now you need to find the area of each small triangle AOC ...

17. Jun 13, 2017

### Helly123

I know that button. I tried before in another thread. But the image wouldnt show up

18. Jun 13, 2017

### Helly123

Ok, for the area of AOC = 1/2 r.r.sin 2b
Area CAB = 1/2 r.r sin 2a
Area BCA = 1/2 r.r sin 2c
So the entire triangle is 1/2. r^2 ( sin 2a + sin 2b + sin 2c)

3) .... AB = 90degrees.1cm ?
BC = 150 degrees.1cm?
AC = 120 degrees.1cm

19. Jun 13, 2017

### Buffu

2 is correct.

In three the angles should be in radians.

20. Jun 13, 2017