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Trigonometry relations

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data


    Straight line segments are drawn from the fixed point P1(0,1) and P2(3,2) to the movable point P, with coordinates (x,0)on the positive x-axis.

    Assuming that 0 ≤ x ≤ 3, show that the angle θ between the two line segments PP1 and PP2 is given by the relation:

    θ= π-arccotx-arccot(3-x/2)



    3. The attempt at a solution


    [tex]tan\theta = (3-0)/(2 - 1) = 3/2 [/tex] For PP1

    [tex]\tan\theta = (3-x)/(2-0)= (3 - x)/2[/tex] For PP2

    [tex]\theta = arctan(3/2) - arctan((3-x)/2)[/tex]

    First of all l do not know if l am on the right path and even if l am on the right path l get stuck here. Please help
     
  2. jcsd
  3. Mar 11, 2009 #2

    tiny-tim

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    Hi Nyasha! :smile:
    No, that's the line from P1 to P2, not P to P1
    I'm not sure either … it's difficult to tell whether you're just guessing …

    to make it clear to the examiner that you're not guessing, give the angles or directions names, so you can make it clear which angles you're subtracting from which, and why …

    in this case, call (0,0) O, and call (3,0) A, and use OPA = π :wink:
     
  4. Mar 11, 2009 #3

    What is OPA =π ?
     
  5. Mar 11, 2009 #4

    tiny-tim

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    uhh? OPA is the angle OPA, from (0,0) to P to (3,0)
     
  6. Mar 11, 2009 #5

    Okay, so any tips on how l prove that θ= π-arccotx-arccot(3-x/2) ? I am really confused/blank on this question.
     
  7. Mar 11, 2009 #6

    tiny-tim

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    just draw it … it's obvious! :smile:
     
  8. Mar 11, 2009 #7
    CCF11032009_00000.jpg


    Tim l have drawn it but it is still not obvious. I still can't figure out how they came up with the relation θ= π-arccotx-arccot(3-x/2)
     
  9. Mar 12, 2009 #8

    tiny-tim

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    waaa!

    oh Nyasha …

    you missed out poor little P! :cry:
     
  10. Mar 12, 2009 #9
    Re: waaa!


    Tim l think l solved it now, l drew a better diagram and then figure out what to do. Here is what l did:


    [tex]\theta= ArcCot((x-0)/(0-1)) = ArcCot(-x) = Pi - ArcCot(x)[/tex] This one is the slope for PP1

    [tex]\theta = ArcCot((3-x)/(2-0) = ArcCot((3-x)/2)[/tex]

    [tex]\theta = Pi - ArcCot(x) - ArcCot((3-x)/2)[/tex] (subtract PP1 from PP2)
     
  11. Mar 12, 2009 #10

    tiny-tim

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    Looks good! :biggrin:
     
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