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Math Help - Trigonometry help?

  1. #16
    4AM
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    Quote Originally Posted by 11rdc11 View Post
    From there either use law of sines or sin or cos to TU
    The laws? I haven't been taught the laws of sines, sin, or cos yet. Is there another way to solve the problem?
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  2. #17
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    Quote Originally Posted by 4AM View Post
    Wait, I don't get why we have to use cos?
    Because the 50 is your Adjacent Side not Hypot as I said (look at the angles to see which side is the hypot)

    Im going to go look at my thread now maybe somones helped me on it take a look if you know statistics and probability.
    http://www.mathhelpforum.com/math-he...tribution.html
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  3. #18
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    Quote Originally Posted by tjsfury View Post
    Because the 50 is your Adjacent Side not Hypot as I said (look at the angles to see which side is the hypot)

    Im going to go look at my thread now maybe somones helped me on it take a look if you know statistics and probability.
    http://www.mathhelpforum.com/math-he...tribution.html
    So the two triangles must be split up separately? I thought the 50 would the opposite side?

    And sorry, I don't know enough about the statistics and probability unit enough to help you with your question. I can hardly understand my current math teacher right now.
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  4. #19
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by 4AM View Post
    So the two triangles must be split up separately? I thought the 50 would the opposite side?

    And sorry, I don't know enough about the statistics and probability unit enough to help you with your question. I can hardly understand my current math teacher right now.
    Yep the triangles have to be split
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  5. #20
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    Hello, 4AM!

    PQ = 50,\;\;\angle PSR = 32^o,\;\;\angle QSR = 216^o.\;\;\text{Find }RS.
    Code:
      -     P *
      :       |   *
      :       |       *
      :     x |           *
      :       |               *
     50       |              32  *
      :     R * - - - - - - - - - - - * S
      :       |            26  *
      :  50-x |           *
      :       |     *
      -     Q *

    Let x \:=\:PR \quad\Rightarrow\quad 50-x\:=\:RQ

    In right triangle PRS\!:\;\;\tan32^o \:=\:\frac{x}{RS} \quad\Rightarrow\quad x \:=\:RS\!\cdot\!\tan32^o\;\;{\color{blue}[1]}

    In right triangle QRS\!:\;\;\tan26^o \:=\:\frac{50-x}{RS} \quad\Rightarrow\quad x \:=\:50-RS\!\cdot\!\tan26^o\;\;{\color{blue}[2]}

    Equate [1] and [2]: . RS\!\cdot\tan32^o \:=\:50 - RS\!\cdot\!\tan26^o \quad\Rightarrow\quad RS\!\cdot\!\tan32^o + RS\!\cdot\!\tan26^o \:=\:50

    Factor: . RS\!\cdot\!(\tan32^o+\tan26^o) \:=\:50 \quad\Rightarrow\quad RS \:=\:\frac{50}{\tan32^o+\tan26^o}

    Therefore: . RS \;\approx\;44.94




    \angle STW = 38.8^o,\;\;\angle STV = 53.8^o,\;\;VW = 12.5 . Find TU.
    Code:
        T * - - - - - - - S
          | * *  38.8
          |   *15*
          |     *     *
          |       *       *
          |         *         *
          |           *           *
          |       53.8 *        38.8 *
          * - - - - - - - * - - - - - - - *
          U               V     12.5     W

    We have: . \angle STW = 38.8^o \quad\Rightarrow\quad \angle TWU = 38.8^o
    Since \angle STV = 58.8^o, then \angle WTV = 15^o\:\text{ and }\:\angle TVU = 53.8^o


    In \Delta TVW, use the Law of Sines:

    . . \frac{TV}{\sin38.8^o} \:=\:\frac{12.5}{\sin15^o} \quad\Rightarrow\quad TV \:\approx\:30.36


    In right triangle TUV\!:\;\;\sin53.8^o \:=\:\frac{TU}{TV} \quad\Rightarrow\quad TU \:=\:30.26\!\cdot\!\sin53.8^o

    Therefore: . TU \;\approx\;24.42

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  6. #21
    4AM
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    Quote Originally Posted by 11rdc11 View Post
    Yep the triangles have to be split
    If they're split though, why is the side still 50 m? Wouldn't it be smaller?
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  7. #22
    Super Member 11rdc11's Avatar
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    Side PQ and QS are both 50
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  8. #23
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    Quote Originally Posted by 11rdc11 View Post
    Side PQ and QS are both 50
    How do you know that though? Sorry, these questions probably seem really stupid to you but I don't really understand how to solve these questions.

    Thank you for your help though! =)
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  9. #24
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by 4AM View Post
    How do you know that though? Sorry, these questions probably seem really stupid to you but I don't really understand how to solve these questions.

    Thank you for your help though! =)
    Because the triangle is an isoceles so two sides are the same
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