A circle is circumscribed around triangle ABC, find length?

[Helly123]

some formula related

I tried to draw the problem

can anyone give me clue how to solve it?

 

[QuantumQuest]

... 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).

 

[Buffu]

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

 

[Helly123]

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

 

[Helly123]

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

can you give me more clue...

 

[Buffu]

can you give me more clue...

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

 

[Helly123]

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

 

[Buffu]

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.

 

[Helly123]

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?

 

[Helly123]

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?

 

[Buffu]

Find theta.

 

[Helly123]

View attachment 205323

Find theta.

I did it https://s21.postimg.org/juzbeyblz/image.jpg

 

[cnh1995]

I did it https://s21.postimg.org/juzbeyblz/image.jpg

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.

 

[Helly123]

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?

 

[cnh1995]

my answer is right? And the formula too?

Yes.

Cant upload through phone) b

I can upload through phone using the upload button below the 'post reply' button.

btw sir,

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

 

[Buffu]

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 ...

 

[Helly123]

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

 

[Helly123]

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

 

[Buffu]

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.

 

[Helly123]

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

 

[Helly123]

Why my answer not match the key answer? :(

 

[Buffu]

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.

 

[Hendrik Boom]

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.

 

[Helly123]

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...

 

[Helly123]

View attachment 205323

Find theta.

Sir, i want to know how to find arc AB BC CA if the triangle like what I draw...

 

[Buffu]

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 ?

 

[Helly123]

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.

 

[Buffu]

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.

 

[Helly123]

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,

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

if I draw this, the length AX and CX no longer R... because the center not in X point

how is it?

 

[Buffu]

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 :-

 

[Helly123]

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 ?

 

[Buffu]

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.

 

[Helly123]

Yes sorry it is angle c.

Your figure should be like this :-
View attachment 205456

Can i say 2phi - theta = twice angle ABC?

 

[Buffu]

Can i say 2phi - theta = twice angle ABC?

Yes.

 

[Kaura]

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