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Math Help - Odd Triangle

  1. #1
    Senior Member DivideBy0's Avatar
    Mar 2007
    Melbourne, Australia

    Odd Triangle

    I have a triangle HPN with two side lengths HP = 54km and HN = 75km and included angle 15 degrees. Let the angles opposite HP and HN be \phi and \theta respectively. By use of the cosine rule I get

    PN=\sqrt{54^2+75^2-2(54)(75) \cos {15}}=26.78km

    Next, I use the sine rule to solve for remaining angles:

    \frac{26.78}{\sin{15}}=\frac{75}{\sin {\theta}}

    \theta=sin^{-1}\left( \frac{75 \sin{15}}{26.78} \right)=46.456 ^ \circ

    The last angle I will find using the sine rule again.

    \frac{26.78}{\sin {15}}=\frac{54}{\sin{\phi}}

    \phi=sin^{-1} \left( \frac{54 \sin 15}{26.78} \right)=31.459^ \circ

    Apprximately, i have a 93 degree triangle here. What did I do wrong?
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    The equation \sin x=a,a\in(0,1) has two solutions in (0,\pi). One solution in \left(0,\frac{\pi}{2}\right) and the other in \left(\frac{\pi}{2},\pi\right).
    You should better use the cosine rule: find cosine of the other two angles using the sides of triangle and then find the angles.
    If \cos x>0\Rightarrow x\in\left(0,\frac{\pi}{2}\right).
    If \cos x<0\Rightarrow x\in\left(\frac{\pi}{2},\pi\right).
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