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anonymous

  • 4 years ago

For the given points P,Q.R, find the approximate measurements of angle PQR. P: (1,-4), Q: (2,7), R: (-2,2)

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  1. Mertsj
    • 4 years ago
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    |dw:1328102812701:dw|

  2. anonymous
    • 4 years ago
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    have to solve it by dot products in calc 3

  3. phi
    • 4 years ago
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    Mertsj has already figured out the lengths of these vectors

  4. anonymous
    • 4 years ago
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    yea but how do you go about to dove this? i have no clue? i am trying all kinds of different things

  5. anonymous
    • 4 years ago
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    solving*

  6. phi
    • 4 years ago
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    First, let me fix it. we want PQR \[(\overrightarrow P -\overrightarrow Q)\cdot (\overrightarrow R -\overrightarrow Q)= |\overrightarrow P -\overrightarrow Q||\overrightarrow R-\overrightarrow Q|\cos\theta\]

  7. phi
    • 4 years ago
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    I assume you know how to find P-Q and R-Q : subtract corresponding elements I'll do P-Q (1,-4) - (2,7)= (1-2, -4-7)= (-1, -11)

  8. phi
    • 4 years ago
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    the length of a vector (x,y) = | (x,y)| = sqrt(x*x + y*y)

  9. phi
    • 4 years ago
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    a dot product of (x,y) and (a,b)= ax+by

  10. phi
    • 4 years ago
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    Did you find R-Q ?

  11. anonymous
    • 4 years ago
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    yea its (-4,-5)

  12. phi
    • 4 years ago
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    Did you find the magnitude of R-Q (and P-Q) ?

  13. anonymous
    • 4 years ago
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    you just square them right?

  14. phi
    • 4 years ago
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    the length of a vector (x,y) = | (x,y)| = sqrt(x*x + y*y) that means you take the first number , square it, square the 2nd number, add together then take the square root

  15. anonymous
    • 4 years ago
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    p-q = sqrt (121) r-q = sqrt (41)

  16. phi
    • 4 years ago
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    double check p-q

  17. anonymous
    • 4 years ago
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    oh its sqrt (122)

  18. phi
    • 4 years ago
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    ok , now the dot product of (p-q) dot (r-q) (-1,-11) dot (-4, -5)

  19. phi
    • 4 years ago
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    dot product of (x,y) and (a,b)= ax+by

  20. anonymous
    • 4 years ago
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    so its (-1*-4 + -11 * -5)

  21. phi
    • 4 years ago
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    yes, but that is just a number=?

  22. anonymous
    • 4 years ago
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    it = 59?

  23. phi
    • 4 years ago
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    yes. now put it all together (p-q) dot (r-q) = |p-q| | r-q| cos A we now everything except A.

  24. anonymous
    • 4 years ago
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    but that will only give us one angle right?

  25. phi
    • 4 years ago
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    This gives us the angle formed by p-q and r-q p-q is a vector with its base at q. same for r-q |dw:1328106371363:dw|

  26. anonymous
    • 4 years ago
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    yea but we need three angles angle P=? , angle Q=? , and angle R =?.

  27. phi
    • 4 years ago
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    find the approximate measurements of angle PQR this means the angle from P to Q to R in other words, Q is the vertex . Just 1 angle

  28. anonymous
    • 4 years ago
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    this is problem 60 in the book. problem 61 is the same question but P: (0, -1, 3), Q = (2,2,1), R= (-2, 2, 4) and answer given in the back is p=78.8 degrees , q= 47.2 degrees and r = 54.0 degree

  29. anonymous
    • 4 years ago
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    thats's why i am confused

  30. phi
    • 4 years ago
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    First, what did you get for angle pqr?

  31. anonymous
    • 4 years ago
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    i can't solve it. sorry i am really bad at it

  32. anonymous
    • 4 years ago
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    so its 59 = sqrt (122) * sqrt (41) cos a

  33. phi
    • 4 years ago
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    the contraction of "it is" is "it's" (sorry , pet peeve!)

  34. phi
    • 4 years ago
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    to find cos a, divide both sides of the equation by (sqrt(122)*sqrt(41))

  35. anonymous
    • 4 years ago
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    lol sorry i am nervous right now

  36. anonymous
    • 4 years ago
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    i get 0.83

  37. phi
    • 4 years ago
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    you can type acos(59/(sqrt(122)*sqrt(41))) in degrees in the google search window to find the angle or use a calculator.

  38. anonymous
    • 4 years ago
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    i get 0.58

  39. phi
    • 4 years ago
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    that's in radians. I would do 2 things. first, use a few more digits in your answer to 59/(sqrt(122)*sqrt(41)) (because your book is finding the answer rounded to tenths of a degree) then use degree mode with the calculator or type in degrees when using google

  40. anonymous
    • 4 years ago
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    oh k i get 33.5 degrees

  41. phi
    • 4 years ago
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    good, matches what I got.

  42. anonymous
    • 4 years ago
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    but thats just one angle. how do i find the other ones

  43. phi
    • 4 years ago
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    If you want the angle with vertex P, which we would name as angle QPR (or angle RPQ), form the vectors (q-p) and (r-p) and do the same thing: (q-p) dot (r-p) = |(q-p)| |(r-p)| cos P

  44. anonymous
    • 4 years ago
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    oh k

  45. anonymous
    • 4 years ago
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    would you help me with other question that i posted?

  46. anonymous
    • 4 years ago
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    and to find the third angle, i could add two angles and make it equal to 180 right?

  47. phi
    • 4 years ago
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    btw, notice that (q-p) = -(p-q), so just negate each element in your (p-q) vector , which you already know. And of course the length of (q-p) = length (p-q)

  48. phi
    • 4 years ago
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    Yes, the third angle = 180 - sum(other two) but it doesn't hurt to do the dot product.

  49. phi
    • 4 years ago
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    as a check, and to get comfortable with the procedure

  50. anonymous
    • 4 years ago
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    yes i need practice.

  51. phi
    • 4 years ago
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    when you post your answers I'll double check them.

  52. anonymous
    • 4 years ago
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    ok

  53. anonymous
    • 4 years ago
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    i get 31.8 degrees

  54. anonymous
    • 4 years ago
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    what would be the setup to find the last angle?

  55. phi
    • 4 years ago
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    yes, looks good. for the last angle, the vertex is R look at how you did the previous two, and look for the pattern. post your setup

  56. anonymous
    • 4 years ago
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    (p-r) * (q-r) = |p-r| |q-r| cos a ?

  57. anonymous
    • 4 years ago
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    i get 114.8 for the last one

  58. phi
    • 4 years ago
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    notice that 180 - (31.8+33.5)= 114.7 which matches your result within rounding error.

  59. anonymous
    • 4 years ago
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    yes

  60. anonymous
    • 4 years ago
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    can you please help me with my other problem?

  61. phi
    • 4 years ago
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    repost it. this one is long enough

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