- #1
Helly123
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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?
Hi Helly123Helly123 said:... can anyone give me clue how to solve it?
the question : (1) how the length of arcs AB, BC, CA, expressed in a, b, c, r.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).
can you give me more clue...Buffu said:Arc-length is ##l = r \theta##.
Helly123 said:can you give me more clue...
maybe likeBuffu said:Can you express arclengths in terms of ##a,b,c ,r ## ?
Helly123 said:maybe like
AB / 2phi R = c / 360 degrees
AB = c/360 . 2 phi R
Maybe the way i draw the circle and triangle is wrong?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.
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.
I did it https://s21.postimg.org/juzbeyblz/image.jpgBuffu said:
Looks good.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?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.
Yes.Helly123 said:my answer is right? And the formula too?
I can upload through phone using the upload button below the 'post reply' button.Helly123 said:Cant upload through phone) b
I'm no sir..just a guy like you, hanging out here!Helly123 said:btw sir,
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?
I know that button. I tried before in another thread. But the image wouldn't show upcnh1995 said:Yes.
I can upload through phone using the upload button below the 'post reply' button.
Ok, for the area of AOC = 1/2 r.r.sin 2bBuffu said:Now you need to find the area of each small triangle AOC ...
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
Helly123 said:Why my answer not match the key answer? :(
I think so...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.
Sir, i want to know how to find arc AB BC CA if the triangle like what I draw...Buffu said:
Helly123 said:Sir, i want to know how to find arc AB BC CA if the triangle like what I draw... View attachment 205419
I meant, totally everything stuck..Buffu said:It is same process.
Where are you stuck ?
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.
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 AXBuffu 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.
Yes sorry it is angle c.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
sorry, but I still have questions regarding that..Buffu said:
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 ?
Can i say 2phi - theta = twice angle ABC?Buffu said:
Helly123 said:Can i say 2phi - theta = twice angle ABC?
A circumscribed circle is a circle that passes through all the vertices of a given polygon, in this case a triangle.
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.
A circumscribed circle is always tangent to each side of the triangle, meaning that it touches each side at exactly one point.
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.
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.