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I'm having a lot of trouble with similar triangles

  1. Jul 26, 2015 #1

    JR Sauerland

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    DMg79Du.png
    Seriously, I understand it sometimes, but when it hits me with something like this, I'm just baffled...
     
  2. jcsd
  3. Jul 26, 2015 #2

    BvU

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    A matter of writing down ##\cos\theta## in two ways: for the left triangle it's ?/50 and for the right one it's 32/? So ##{?\over 50} = {32\over ?}##
     
  4. Jul 26, 2015 #3

    JR Sauerland

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    See, that's the part I get because of SOH CAH TOA. It's cosine, CAH, Adjacent over the Hypotenuse. The other one is completely the opposite. I set the problem up just like you did, but I really don't know where to go from there. ##{?\over 50} = {32\over ?}## That is what is baffling me.
     
  5. Jul 26, 2015 #4

    BvU

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    Multiply left and right with 50 ?, to get ##?^2 = 50 \times 32##
     
  6. Jul 26, 2015 #5

    JR Sauerland

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    Perfect! Now I completely understand the logic behind it! On another note, ##{30\over ?} = {?\over 32}##, if it had been formatted like this, do I simply take the reciprocal like this: ##{?\over 30} = {32\over ?}##
    Basically, I'm getting at the fact that the variable (which typically is x) isn't supposed to be on the bottom?
     
  7. Jul 26, 2015 #6

    Mentallic

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    Do you understand how BvU went from

    [tex]\frac{x}{50}=\frac{32}{x}[/tex]
    to
    [tex]x^2=50\times 32[/tex]
     
  8. Jul 26, 2015 #7

    JR Sauerland

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    I think so. Didn't he multiply both sides by (50x)? It eliminated x in the denominator of the right side, and squared the x in the left side due to the fact that it was on top. I've heard you could cross multiple too though, although I don't know if it would be completely different or not...
     
  9. Jul 26, 2015 #8

    Mentallic

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    Right, so apply the same idea to your new problem. Taking the reciprocal of both sides is a valid operation, but it doesn't help you because you want to remove the variable from the denominator, which just so happens to give us a quadratic (squared value of x) in this sort of problem.

    Also, cross multiplying is exactly multiplying both sides by 50x. It's just a means to help students visualize or to rote learn the process more easily.
     
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