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

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.

2. ## Triangle trigonometry

Hello Tovi

Welcome to Math Help Forum!
Originally Posted by Tovi
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

3. Nice explanation Grandad.

I assume that the line CG is perpendicular to AB

That simplifies things considerably, but it is solvable otherwise.

4. Hello aidan
Originally Posted by aidan
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