Results 1 to 4 of 4

Thread: I can't figure out this triangle-based problem

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    1

    I can't figure out this triangle-based problem

    I have included a drawing as an attachment.

    The following lengths are given: c, d, H and L

    The angle indicated with green is Theta. This is given too.

    The two red markings indicate 90 degree angles.

    I need to find length "a".

    I'm really stuck on this one. It's probably not that hard, but I could use some clues as it's only a small part of a bigger physics problem.
    Attached Thumbnails Attached Thumbnails I can't figure out this triangle-based problem-triangles.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Triangle trigonometry

    Hello Tovi

    Welcome to Math Help Forum!
    Quote Originally Posted by Tovi View Post
    I have included a drawing as an attachment.

    The following lengths are given: c, d, H and L

    The angle indicated with green is Theta. This is given too.

    The two red markings indicate 90 degree angles.

    I need to find length "a".

    I'm really stuck on this one. It's probably not that hard, but I could use some clues as it's only a small part of a bigger physics problem.
    I've added some detail to your drawing. See attachment.

    I assume that the line $\displaystyle CG$ is perpendicular to $\displaystyle AB$. In which case $\displaystyle \angle CDF = \angle CBG = \phi$, say.

    Then $\displaystyle \tan \theta = \frac{h}{c}$ and $\displaystyle \tan\phi = \frac{h}{d}$

    And $\displaystyle CD = \frac{a}{\tan\theta}=\frac{ac}{h} = \frac{L-a}{\tan\phi}=\frac{d(L-a)}{h}$

    $\displaystyle \Rightarrow ac = d(L-a)=dL-da$

    $\displaystyle \Rightarrow a = \frac{dL}{c+d}$

    Looks pretty simple when you see it, doesn't it?

    Grandad
    Attached Thumbnails Attached Thumbnails I can't figure out this triangle-based problem-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jan 2009
    Posts
    591

    Thumbs up

    Nice explanation Grandad.

    I assume that the line CG is perpendicular to AB

    That simplifies things considerably, but it is solvable otherwise.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello aidan
    Quote Originally Posted by aidan View Post
    Nice explanation Grandad.


    That simplifies things considerably, but it is solvable otherwise.
    In fact I was wrong in saying 'I assume...' because we can deduce that $\displaystyle CG$ and $\displaystyle AB$ are perpendicular, since $\displaystyle \angle EDC = \angle CAG = \theta$ (given), and $\displaystyle ED$ and $\displaystyle AC$ are given perpendicular. So by angle sum of triangle (or indeed the quadrilateral) $\displaystyle \angle AGD = 90^o$.

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 21st 2010, 08:46 AM
  2. Replies: 2
    Last Post: Jun 4th 2010, 07:02 AM
  3. Replies: 1
    Last Post: Apr 30th 2009, 01:28 AM
  4. need help with differential (problem based)
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Sep 6th 2008, 05:03 AM
  5. Help me figure out area of triangle!!
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: May 27th 2007, 08:36 AM

Search Tags


/mathhelpforum @mathhelpforum