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Finding the height of this triangle

  1. Jan 11, 2007 #1
    The problem is to find h on this triangle:


    With the help of the law of sines I've already finished this problem. But I tried doing it a different way and my new solution isn't working and I can't figure out why. Here's what I did:

    (1) Break the bottom part like so


    (2) Use tan to solve for x in two ways

    tan40 = h/x (so x = (tan40)/h)
    tan47 = h/(125-x) (so x = 125 - hcot47)

    (3) Since both those equations equal x, they equal each other, so I do this:

    (tan40)/h = 125 - hcot47

    Multiplying both sides by h, blah blah blah, I get this

    h2cot47 - 125h + tan40 = 0.

    So I have a quadratic equation in h. When I use the quadratic formula and do all the solving (which I'll omit since it would take a lot of space to write) I get an incorrect answer. Where did I go wrong?
  2. jcsd
  3. Jan 11, 2007 #2
    Your mistake is here: tan40 = h/x (so x = (tan40)/h)

    x should be equal to h / tan40
  4. Jan 11, 2007 #3


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    Staff Emeritus
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    Gold Member

    You correctly say

    [tex] \tan 40 = \frac h x [/tex]

    try solving this for x again. Perhaps you will find a different result. :smile:
  5. Jan 11, 2007 #4
    LOL!!!! I combed over my solution about a thousand times looking for what I did wrong and I couldn't see that. Thanks guys.
  6. Jan 13, 2007 #5
    The way I would of done it would of been like so...

    I wouldn't of broken up 125 to x and 125-x.. I would of found one of the other sides using sin law because we can figure out the missing angle

    [tex]\Theta = 180 - (40+47)[/tex]

    [tex]\Theta = 93\cdot[/tex]

    then apply sin law

    [tex]\frac{125}{sin93} = \frac{x}{sin40}[/tex]

    [tex]125*sin40 = x*sin93[/tex]

    [tex]\frac{125*sin40}{sin93} = x[/tex]

    [tex]80.5 = x[/tex]

    then you have the right triangle which is a right angel triangle with the hypotinuse and a given angel.. thats all you need to find h using
    [tex]sin47 = \frac {h}{80.5}[/tex]
  7. Jan 13, 2007 #6

    Gib Z

    User Avatar
    Homework Helper

    No, he said he already did it that way, But did it another way and had a problem with that. Nice bit of tex though.
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