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A circle is circumscribed around triangle ABC, find length?

  1. Jun 12, 2017 #1
    • New poster has been reminded to use the Homework Help Template when starting threads in the schoolwork forums
    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?
     
  2. jcsd
  3. Jun 12, 2017 #2

    QuantumQuest

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    Gold Member

    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).
     
  4. Jun 12, 2017 #3
    Arc-length is ##l = r \theta##.
     
  5. Jun 12, 2017 #4
    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
     
  6. Jun 12, 2017 #5
    can you give me more clue.....
     
  7. Jun 12, 2017 #6
    Can you express arclengths in terms of ##a,b,c ,r ## ?
     
  8. Jun 12, 2017 #7
    maybe like
    AB / 2phi R = c / 360 degrees
    AB = c/360 . 2 phi R
     
  9. Jun 12, 2017 #8
    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.
     
  10. Jun 12, 2017 #9
    Maybe the way i draw the circle and triangle is wrong?
     
  11. Jun 13, 2017 #10
    That's, the first time i saw it too...

    " can you find the angle subtended by each side at centre ? " how to find that?
     
  12. Jun 13, 2017 #11
    g4220-6-2-28asd.png

    Find theta.
     
  13. Jun 13, 2017 #12
  14. Jun 13, 2017 #13

    cnh1995

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    Homework Helper

    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.
     
  15. Jun 13, 2017 #14
    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?
     
  16. Jun 13, 2017 #15

    cnh1995

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    Homework Helper

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

    I'm no sir..just a guy like you, hanging out here!
     
  17. Jun 13, 2017 #16
    Now you need to find the area of each small triangle AOC ...
     
  18. Jun 13, 2017 #17
    I know that button. I tried before in another thread. But the image wouldnt show up
     
  19. Jun 13, 2017 #18
    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
     
  20. Jun 13, 2017 #19
    2 is correct.

    In three the angles should be in radians.
     
  21. Jun 13, 2017 #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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