Results 1 to 2 of 2

Thread: Odd Triangle

  1. #1
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432

    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 $\displaystyle \phi$ and $\displaystyle \theta$ respectively. By use of the cosine rule I get

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

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

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

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

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

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

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

    Apprximately, i have a 93 degree triangle here. What did I do wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    The equation $\displaystyle \sin x=a,a\in(0,1)$ has two solutions in $\displaystyle (0,\pi)$. One solution in $\displaystyle \left(0,\frac{\pi}{2}\right)$ and the other in $\displaystyle \left(\frac{\pi}{2},\pi\right)$.
    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 $\displaystyle \cos x>0\Rightarrow x\in\left(0,\frac{\pi}{2}\right)$.
    If $\displaystyle \cos x<0\Rightarrow x\in\left(\frac{\pi}{2},\pi\right)$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Apr 23rd 2011, 08:10 AM
  2. Replies: 3
    Last Post: Apr 30th 2009, 07:41 AM
  3. Replies: 1
    Last Post: Oct 28th 2008, 07:02 PM
  4. Replies: 7
    Last Post: Jul 19th 2008, 06:53 AM
  5. Replies: 27
    Last Post: Apr 27th 2008, 10:36 AM

Search Tags


/mathhelpforum @mathhelpforum