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

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

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

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

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

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

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