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Math Help - Find H on this triangle

  1. #1
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    Find H on this triangle

    here's the pic I made... (find H)


    now, I'm confused on how to set it up, should it be

    tan(18) = h / 5+x
    tan(32) = h / x

    or

    tan(18) = h / b
    tan(32) = h / b-5

    (b = base)

    or neither?

    please tell me which set-up it is and why, i can't get it
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Arez View Post
    here's the pic I made... (find H)


    now, I'm confused on how to set it up, should it be

    tan(18) = h / 5+x
    tan(32) = h / x

    or

    tan(18) = h / b
    tan(32) = h / b-5

    (b = base)

    or neither?

    please tell me which set-up it is and why, i can't get it
    Both are correct, except that you should use parentheses for the denominators so that
    h/ 5+x would appear h / (5+x)
    h / b -5 would appear h /(b-5)

    Nothing to be confused about.
    tan is opposite side / adjacent side .......that's it.
    No matter how you come up with the measurements of the sides, as long as you carefully follow tan = opp/adj, then you're okay.

    ---
    EDIT:

    Opsss, I'm wrong!
    After reading masters' reply (which is correct), I am wrong with mine.
    tan(32 deg) cannot be h /(5 +x) nor h /(b -5)....because h is not the opposite side of 32 degrees.

    Must be that I just woke up.
    Still, I shouldn't have missed that tan(32 deg)....

    Sometimes, though, I like being wrong. :-)
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Arez View Post
    here's the pic I made... (find H)


    now, I'm confused on how to set it up, should it be

    tan(18) = h / 5+x
    tan(32) = h / x

    or

    tan(18) = h / b
    tan(32) = h / b-5

    (b = base)

    or neither?

    please tell me which set-up it is and why, i can't get it
    Not enough info, I fear. According to your diagram:

    \tan 18=\frac{H}{5+X} is correct!

    The 32 degree angle has no opposite side defined. H is the opposite side of the 18 degree angle. So,

    \tan 32 \neq \frac{H}{X}
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by ticbol View Post
    Opsss, I'm wrong!
    After reading masters' reply (which is correct), I am wrong with mine.

    Sometimes, though, I like being wrong. :-)
    It keeps us humble, you know.
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