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Math Help - Four segments, find angle?

  1. #1
    Newbie sjcomp's Avatar
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    Four segments, find angle?

    Given the geometry shown below I need to find angle r. I now all the sides (a,b,c,d). I do NOT know what angle K is. Here is the geometry setup:

    I approached this problem the following way:
    EC = AB + 2*BC*sin(k)
    DE^2 = DC^2 + EC^2 - 2*DC*EC*cos(r)
    I have two equations and three unknowns. Which means I am missing one more equation. Any suggestions? Or maybe a better approach?
    Thanks.
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  2. #2
    Senior Member JaneBennet's Avatar
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    Let the H be the foot of the perpendicular from D to the line AB produced on the side of A, and let I be the foot of the perpendicular from C to the line AB produced on the side of B. Then the angles HDA and ICB are equal to κ, and HI = b\sin{\kappa}+d+a\sin{\kappa}=c\cos{r}.

    \therefore\ \sin{\kappa}\ =\ \frac{c\cos{r}-d}{b+a}\quad\ldots\fbox{1}

    Also if DH meets the line EC produced on the side of E at J, then DJ =
    (b-a)\cos{\kappa}=c\sin{r}.

    \therefore\ \cos{\kappa}\ =\ \frac{c\sin{r}}{b-a}\quad\ldots\fbox{2}

    From [1] and [2], κ can be eliminated.
    Last edited by JaneBennet; December 22nd 2007 at 12:33 PM. Reason: Minor LaTeX formatting
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  3. #3
    Newbie sjcomp's Avatar
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    Thanks a lot JaneBennet. I see what I was missing there
    Quote Originally Posted by JaneBennet View Post
    From [1] and [2], κ can be eliminated.
    I used this formula to connect cos(k) and sin(k):
    <br />
\cos{\kappa}\ =\ \sqrt{1- \sin{\kappa}^2}  \quad\ldots\fbox{3}<br />
    After substitution I get a formula that has \sin{r}^2, \cos{r}^2 and sin{r}. This makes it tricky to get actual r. Any suggestions?
    Last edited by sjcomp; December 22nd 2007 at 02:42 PM. Reason: Correcting formula
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  4. #4
    Newbie sjcomp's Avatar
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    After playing with the equations I boiled it down to this formula:
    2abc^2(\sin{r}^2-\cos{r}^2)-2d\cos{r}(b-a)^2 = -(b^2c^2+a^2c^2+d^2(b-a)^2) \quad\ldots\fbox{4} .
    Everything on the right side is a constant. But how do I convert the difference of squares there? It would be convenient if am able to collapse it to \cos{r}^2, but how?
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  5. #5
    Senior Member JaneBennet's Avatar
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    Itís very complicated, I know. But if you do everything carefully, you should end up with a quadratic equation in \cos{r}, which can always be solved using the quadratic formula.
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  6. #6
    Newbie sjcomp's Avatar
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    More calculations

    It's more interesting than I thought I tried another approach for substitution, which lead me pretty much to the same result, meaning I still have \sin{r}^2-\cos{r}^2. Here are my calculations. First I simplify both sides of the equations and then I bring them together. I checked the signs a few times and they all match properly. Any suggestions?

    Thanks a lot.
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  7. #7
    Newbie sjcomp's Avatar
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    I did find the solution with the help of Mathcad. Thanks a lot for the help.
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