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Mimi_x3

  • 3 years ago

Geometry Question.

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

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

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

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

  6. Mimi_x3
    • 3 years ago
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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?

  7. lalaly
    • 3 years ago
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    since*

  8. Mimi_x3
    • 3 years ago
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    Hm,okay; I will try from that..

  9. Mimi_x3
    • 3 years ago
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    thanks anyway

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

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

  12. Mimi_x3
    • 3 years ago
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    Nope, only the first one.

  13. lalaly
    • 3 years ago
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    ok:)

  14. inkyvoyd
    • 3 years ago
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    That's a stupid question.

  15. inkyvoyd
    • 3 years ago
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    * problem

  16. razor99
    • 3 years ago
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    may i come here

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

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

  19. inkyvoyd
    • 3 years ago
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    |dw:1337940741536:dw|

  20. Mimi_x3
    • 3 years ago
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    lol, please delete that inky..

  21. inkyvoyd
    • 3 years ago
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    |dw:1337940823123:dw|

  22. inkyvoyd
    • 3 years ago
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    That's x.

  23. inkyvoyd
    • 3 years ago
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    I understand the problem now.

  24. inkyvoyd
    • 3 years ago
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    |dw:1337940877496:dw|

  25. inkyvoyd
    • 3 years ago
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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.

  26. inkyvoyd
    • 3 years ago
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    @Mimi_x3

  27. inkyvoyd
    • 3 years ago
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    |dw:1337941268446:dw|

  28. inkyvoyd
    • 3 years ago
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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

  29. inkyvoyd
    • 3 years ago
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    |dw:1337941693329:dw|

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

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

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

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

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

  35. inkyvoyd
    • 3 years ago
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    http://www.twiddla.com/847714

  36. Mimi_x3
    • 3 years ago
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    I'm lost..

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

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

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

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

  41. lalaly
    • 3 years ago
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    and TO=x+a

  42. lalaly
    • 3 years ago
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    did u get it @Mimi_x3

  43. Mimi_x3
    • 3 years ago
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    yeah; thank you!

  44. lalaly
    • 3 years ago
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    :D

  45. inkyvoyd
    • 3 years ago
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    @Mimi_x3 , rage

  46. .Sam.
    • 3 years ago
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    I was using angle JOB lol

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