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Mimi_x3 Group Title

Geometry Question.

  • 2 years ago
  • 2 years ago

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  1. Mimi_x3 Group Title
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    • 2 years ago
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  2. Mimi_x3 Group Title
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    i don't know how to start..can someone please help..

    • 2 years ago
  3. lalaly Group Title
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    i am still trying to do it but something is confusing me whats the x?

    • 2 years ago
  4. Mimi_x3 Group Title
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    I don't know; I'm confused as well, where did the x come from? lol

    • 2 years ago
  5. lalaly Group Title
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    lol i did half of the work but got to an equatoin with x in it ...

    • 2 years ago
  6. Mimi_x3 Group Title
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    Well, I don't know how to do it; where the x come from? Can you give me a hint how to start?

    • 2 years ago
  7. lalaly Group Title
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    since*

    • 2 years ago
  8. Mimi_x3 Group Title
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    Hm,okay; I will try from that..

    • 2 years ago
  9. Mimi_x3 Group Title
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    thanks anyway

    • 2 years ago
  10. lalaly Group Title
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    sorry:( .. im sure the other people viewing would have an idea

    • 2 years ago
  11. lalaly Group Title
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    i can do part b xD do u need help with that?

    • 2 years ago
  12. Mimi_x3 Group Title
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    Nope, only the first one.

    • 2 years ago
  13. lalaly Group Title
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    ok:)

    • 2 years ago
  14. inkyvoyd Group Title
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    That's a stupid question.

    • 2 years ago
  15. inkyvoyd Group Title
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    * problem

    • 2 years ago
  16. razor99 Group Title
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    may i come here

    • 2 years ago
  17. .Sam. Group Title
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    where the 'x' come from in equation 'a'?

    • 2 years ago
  18. Mimi_x3 Group Title
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    i don't know that is why I'm lost; or probably there's a mistake :/

    • 2 years ago
  19. inkyvoyd Group Title
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    |dw:1337940741536:dw|

    • 2 years ago
  20. Mimi_x3 Group Title
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    lol, please delete that inky..

    • 2 years ago
  21. inkyvoyd Group Title
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    |dw:1337940823123:dw|

    • 2 years ago
  22. inkyvoyd Group Title
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    That's x.

    • 2 years ago
  23. inkyvoyd Group Title
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    I understand the problem now.

    • 2 years ago
  24. inkyvoyd Group Title
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    |dw:1337940877496:dw|

    • 2 years ago
  25. inkyvoyd Group Title
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    As to how I figured it out, I began by constructing that perpendicular line segment, and then I assumed the statement was true. I found that sin(theta-phi)=sin(angle b), and a sin(b)= that line segment I constructed. Then, sin(phi)=opp/hyp=contructed line segment/(a+something), or (a+something)sin(phi) =a sin(theta-phi) and the only value for something that makes sense is that segment.

    • 2 years ago
  26. inkyvoyd Group Title
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    @Mimi_x3

    • 2 years ago
  27. inkyvoyd Group Title
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    |dw:1337941268446:dw|

    • 2 years ago
  28. inkyvoyd Group Title
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    |dw:1337941445676:dw| a sin(theta-phi)=constructed line segment -> let's call that segment length y Then, (a+x) sin(phi)=opp/hyp ->opp is segment y, and hyp would be x+a - the only definition of x that works is in my drawing. Thus, (a+x)sin(phi)=(a+x)*(opp/hyp)=(a+x)(y)/(x+a)=y and a sin(theta-phi) is also y, and so we can substitute (y)-(y)=0

    • 2 years ago
  29. inkyvoyd Group Title
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    |dw:1337941693329:dw|

    • 2 years ago
  30. inkyvoyd Group Title
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    @lalaly , do you at least understand my explanation?

    • 2 years ago
  31. .Sam. Group Title
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    Max I can get is \[asin(\theta)-(x+a)sin(\phi)=0\] assuming x was there |dw:1337941920722:dw|

    • 2 years ago
  32. inkyvoyd Group Title
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    @.Sam. , that's what I meant... I'm sure that that is x

    • 2 years ago
  33. Mimi_x3 Group Title
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    First, where did sine come from? I havent done geometry in a while..

    • 2 years ago
  34. .Sam. Group Title
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    I've ended up with another solution \[rcos(90+\phi-\theta)-(x+r)sin(\phi)=0\]

    • 2 years ago
  35. inkyvoyd Group Title
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    http://www.twiddla.com/847714

    • 2 years ago
  36. Mimi_x3 Group Title
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    I'm lost..

    • 2 years ago
  37. lalaly Group Title
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    angle TBO= 180-(180-theta+phi)=theta-phi \[\frac{(x+a)}{\sin(\theta-\phi)}=\frac{a}{\sin \phi}\]\[(x+a)\sin \phi=asin(\theta-\phi)\]

    • 2 years ago
  38. lalaly Group Title
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    so\[asin(\theta-\phi)-(x+a)\sin \phi=0\]

    • 2 years ago
  39. lalaly Group Title
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    i assumed that the part of TC that is outside the circle is x ... and it worked!! :D

    • 2 years ago
  40. lalaly Group Title
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    and then i used the sin rule after finding TBO O is the center

    • 2 years ago
  41. lalaly Group Title
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    and TO=x+a

    • 2 years ago
  42. lalaly Group Title
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    did u get it @Mimi_x3

    • 2 years ago
  43. Mimi_x3 Group Title
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    yeah; thank you!

    • 2 years ago
  44. lalaly Group Title
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    :D

    • 2 years ago
  45. inkyvoyd Group Title
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    @Mimi_x3 , rage

    • 2 years ago
  46. .Sam. Group Title
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    I was using angle JOB lol

    • 2 years ago
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