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Math Help - Right Triangle Trig. ?

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Unhappy Right Triangle Trig. ?

    i dont know how to do these that involve using trig identities to transform one side to the other...
    1. (1+ cos [theta]) (1-cos [theta]) = sin^2 [theta].

    2. sin [theta]/ cos [theta] + cos [theta]/ sin [theta] =csc [theta] sec [theta]

    3. Tan [theta] + Cot [theta]/ tan [theta] = csc^2 [theta]

    then there is one i dont get...i can draw the diagram...but the rest i dont get

    4. a 6-ft. person walks from the base of a sreetlight directly toward the tip of the shadow cast by the streetlight. when the person is 16 ft from it and 5 ft from the tip of the streetlight shadow, the person's shadow starts to appear beyond the streetlight's shadow.


    A) use a trig function to write an equation involving the unknown quantity.

    B) what is the height of the streetlight...


    i know i am asking for a lot...any help would be great, especially for 4, but any help will be greatly appreciated...thank you
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  2. #2
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    I think I understood your problem, the key here is that you are given a similar triangle, the 6ft person when he is 5 ft from the tip of the shadow.. that's your 2nd triangle.

    Use that one to get either the angle for the triangle (which will be the same angle for the triangle formed by the streetlight and the long shadow), or simply use the similar triangles rule and solve for the height of the streetlight.

    Let me know if I interpreted your problem correctly.
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  3. #3
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    let me know if this is what your problem looks like...
    as you see similar triangles or using trigonometry will give you the same answer

    sorry for the poor quality of the drawing:






    For the other problems, 1 is identities or something like that, I don't quite remember, anyways, multiply and you'll get 1-cos^2, which by the law is equal to sin^2.
    The others are also based on these identities that I'm talking about, although I'm not sure if that's what they're called, it's been many years since I saw that in school, so it's just a matter of playing with them until you get your answer.
    Last edited by enmascarado; April 2nd 2008 at 07:05 PM. Reason: image attached
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  4. #4
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    1. LHS =  (1 + cos \theta)(1-cos \theta)

    This is the difference of two squares, so

    = (1-cos^2 \theta)
    Then use the trigonometric identity cos^2\theta + sin^2\theta = 1

    =  1- (1-sin^2\theta)<br />
           = sin^2\theta<br />
           = RHS

    2.
    tan\theta is also \frac{sin\theta}{cos\theta} . Inverse tan, (not arctan) is cot \theta.

    cot\theta = \frac{1}{tan\theta}<br />
= \frac{cos\theta}{sin\theta}

    Using this you should be able to solve 2.

    3.for this question, I'm pretty sure you mean:

    \frac{tan\theta + cot\theta}{tan\theta}
    = 1 + (cot\theta)( \frac{1}{tan\theta})

    From question two, I've explained that \frac{1}{tan\theta} = cot\theta
    In this case, use the identity 1 + cot^2\theta = cosec^2\theta
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  5. #5
    Member >_<SHY_GUY>_<'s Avatar
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    what is LHS?

    Quote Originally Posted by Gusbob View Post
    1. LHS =  (1 + cos \theta)(1-cos \theta)

    This is the difference of two squares, so

    = (1-cos^2 \theta)
    Then use the trigonometric identity cos^2\theta + sin^2\theta = 1

    =  1- (1-sin^2\theta)<br />
           = sin^2\theta<br />
           = RHS

    2.
    tan\theta is also \frac{sin\theta}{cos\theta} . Inverse tan, (not arctan) is cot \theta.

    cot\theta = \frac{1}{tan\theta}<br />
= \frac{cos\theta}{sin\theta}

    Using this you should be able to solve 2.

    3.for this question, I'm pretty sure you mean:

    \frac{tan\theta + cot\theta}{tan\theta}
    = 1 + (cot\theta)( \frac{1}{tan\theta})

    From question two, I've explained that \frac{1}{tan\theta} = cot\theta
    In this case, use the identity 1 + cot^2\theta = cosec^2\theta
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  6. #6
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    LHS is left hand side.

    RHS is right hand side.
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  7. #7
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by Gusbob View Post
    LHS is left hand side.

    RHS is right hand side.
    Thanks...
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