A circle is circumscribed around triangle ABC, find length?

In summary: The key answer is root 2, (root 6 + root 2)/2 , root 3In summary, the problem asks for the length of the sides of a triangle, not the length of the arcs.
  • #1
Helly123
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IMG_0936_1.jpg


some formula related
nratbig.gif


I tried to draw the problem
Untitled.png


can anyone give me clue how to solve it?
 
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  • #2
Helly123 said:
... can anyone give me clue how to solve it?
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
Arc-length is ##l = r \theta##.
 
  • #4
QuantumQuest said:
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).
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
upload_2017-6-13_6-48-14.png
 
  • #5
Buffu said:
Arc-length is ##l = r \theta##.
can you give me more clue...
 
  • #6
Helly123 said:
can you give me more clue...

Can you express arclengths in terms of ##a,b,c ,r ## ?
 
  • #7
Buffu said:
Can you express arclengths in terms of ##a,b,c ,r ## ?
maybe like
AB / 2phi R = c / 360 degrees
AB = c/360 . 2 phi R
 
  • #8
Helly123 said:
maybe like
AB / 2phi R = c / 360 degrees
AB = c/360 . 2 phi R

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
Buffu said:
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.
Maybe the way i draw the circle and triangle is wrong?
 
  • #10
Buffu said:
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.

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
g4220-6-2-28asd.png


Find theta.
 
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  • #13
Helly123 said:
Looks good.

Your image link contains a lot of ads and the actual image might take some time to load.
Next time onwards, use the upload button to directly add the images in the post.
 
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  • #14
cnh1995 said:
Looks good.

Your image link contains a lot of ads and the actual image might take some time to load.
Next time onwards, use the upload button to directly add the images in the post.
Ok thanks, I am using phone so i sent link. Usually i upload. (Cant upload through phone) btw sir, my answer is right? And the formula too?
 
  • #15
Helly123 said:
my answer is right? And the formula too?
Yes.
Helly123 said:
Cant upload through phone) b
I can upload through phone using the upload button below the 'post reply' button.

Helly123 said:
btw sir,
I'm no sir..just a guy like you, hanging out here!
 
  • #16
Helly123 said:
Ok thanks, I am using phone so i sent link. Usually i upload. (Cant upload through phone) btw sir, my answer is right? And the formula too?

Now you need to find the area of each small triangle AOC ...
 
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  • #17
cnh1995 said:
Yes.

I can upload through phone using the upload button below the 'post reply' button.
I know that button. I tried before in another thread. But the image wouldn't show up
 
  • #18
Buffu said:
Now you need to find the area of each small triangle AOC ...
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
Helly123 said:
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

2 is correct.

In three the angles should be in radians.
 
  • #20
AC = 2/3 phi rad
BC = 5/6 phi rad
AB = 1/2 phi rad

The key answer is root 2, (root 6 + root 2)/2 , root 3
 
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  • #21
Why my answer not match the key answer? :(
 
  • #22
Helly123 said:
Why my answer not match the key answer? :(

The question wants length of sides of the triangle not the length of the arcs which you calculated.
 
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  • #23
The boxes with 2-1, 2-2, 2-3, etc in them seem to number the subquestions. Kind of where you would fill in the answers to question 2-1, 2-2, 2-3, etc. Except you probably have to fill them in on a separate answer sheet.
 
  • #24
Hendrik Boom said:
The boxes with 2-1, 2-2, 2-3, etc in them seem to number the subquestions. Kind of where you would fill in the answers to question 2-1, 2-2, 2-3, etc. Except you probably have to fill them in on a separate answer sheet.
I think so...
 
  • #25
Buffu said:
Sir, i want to know how to find arc AB BC CA if the triangle like what I draw...
untitled-png.png
 
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  • #26
Helly123 said:
Sir, i want to know how to find arc AB BC CA if the triangle like what I draw... View attachment 205419

It is same process.
Where are you stuck ?
 
  • #27
Buffu said:
It is same process.
Where are you stuck ?
I meant, totally everything stuck..
what's the correlation of let's say angle ABC to R, since ABC not on the same center as circle center?
and L = tetha . R cannot be applied anymore. since the R not in the triangle.
 
  • #28
Helly123 said:
I meant, totally everything stuck..
what's the correlation of let's say angle ABC to R, since ABC not on the same center as circle center?
and L = tetha . R cannot be applied anymore. since the R not in the triangle.

Ok first try with angle subtended on the centre by the line AB.

Also ##l = r\theta## is applicable here since it does not matter whether it is in te triangle or not. It is true by the definition of angle in radians.

I think what you are confused with is the fact that if angle b is half the angle subtended at the centre by the line AB or not. It is also true.
 
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  • #29
Buffu said:
Ok first try with angle subtended on the centre by the line AB.

Also ##l = r\theta## is applicable here since it does not matter whether it is in te triangle or not. It is true by the definition of angle in radians.

I think what you are confused with is the fact that if angle b is half the angle subtended at the centre by the line AB or not. It is also true.

the angle substended by line AB is the c angle?

And even though the triangle is like what i have, the tetha substended by line AB twice angle c,
Buffu said:
Ok first try with angle subtended on the centre by the line AB.

Also ##l = r\theta## is applicable here since it does not matter whether it is in te triangle or not. It is true by the definition of angle in radians.

I think what you are confused with is the fact that if angle b is half the angle subtended at the centre by the line AB or not. It is also true.
if I draw tetha in my triangle like this, and want to find arc AC, but ended up finding arc AX
2_-_2015_mat_a.png
if I draw this, the length AX and CX no longer R... because the center not in X point

how is it?
2_-_2015_mat_a_2.png
 
  • #30
Helly123 said:
the angle substended by line AB is the c angle?

And even though the triangle is like what i have, the tetha substended by line AB twice angle c,
if I draw tetha in my triangle like this, and want to find arc AC, but ended up finding arc AX
View attachment 205447if I draw this, the length AX and CX no longer R... because the center not in X point

how is it?
View attachment 205448
Yes sorry it is angle c.

Your figure should be like this :-
g4220-6-2-2dsf8.png
 
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  • #31
Buffu said:
Yes sorry it is angle c.

Your figure should be like this :-
View attachment 205456
sorry, but I still have questions regarding that..
the tetha times R is to find Arc AC right?
while tetha is twice angle ABC ?
 
  • #32
Helly123 said:
sorry, but I still have questions regarding that..
the tetha times R is to find Arc AC right?
while tetha is twice angle ABC ?

For ##2\pi - \theta## not ##\theta## I drew theta other way because it was difficult to draw on the correct side.
 
  • #33
Buffu said:
Yes sorry it is angle c.

Your figure should be like this :-
View attachment 205456
Can i say 2phi - theta = twice angle ABC?
 
  • #34
Helly123 said:
Can i say 2phi - theta = twice angle ABC?

Yes.
 
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  • #35
Try to use the previous arc length equation and the law of cosines to find different arc length segments of the circle and work from there
 
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Related to A circle is circumscribed around triangle ABC, find length?

1. What is the definition of a circumscribed circle?

A circumscribed circle is a circle that passes through all the vertices of a given polygon, in this case a triangle.

2. How is the length of a circumscribed circle determined?

The length of a circumscribed circle is determined by the distance between the center of the circle and any of the vertices of the triangle.

3. What is the relationship between a circumscribed circle and a triangle?

A circumscribed circle is always tangent to each side of the triangle, meaning that it touches each side at exactly one point.

4. How do you calculate the length of a circumscribed circle?

The length of a circumscribed circle can be calculated using the formula C = 2πr, where C is the circumference of the circle and r is the radius. The radius can be found by dividing the length of any side of the triangle by 2sin(θ), where θ is the angle opposite the side.

5. Can a triangle have more than one circumscribed circle?

No, a triangle can only have one circumscribed circle. This is because the center of the circle is determined by the intersection of the perpendicular bisectors of the sides of the triangle, and there can only be one point that satisfies this condition.

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