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Trigonometry relations

  • Thread starter Nyasha
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1. Homework Statement


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
 

tiny-tim

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Hi Nyasha! :smile:
[tex]tan\theta = (3-0)/(2 - 1) = 3/2 [/tex] For PP1
No, that's the line from P1 to P2, not P to P1
First of all l do not know if l am on the right path
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:
 
127
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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:

What is OPA =π ?
 

tiny-tim

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uhh? OPA is the angle OPA, from (0,0) to P to (3,0)

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

tiny-tim

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just draw it … it's obvious! :smile:
 
127
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just draw it … it's obvious! :smile:
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)
 

tiny-tim

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waaa!

oh Nyasha …

you missed out poor little P! :cry:
 
127
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Re: waaa!

oh Nyasha …

you missed out poor little P! :cry:

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)
 

tiny-tim

Science Advisor
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Looks good! :biggrin:
 

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