Trigonometrical Problem

Question

I think this is a bit odd but I am juggling since hours with $\sin$, $\cos$, $\tan$ and other stuff to proof a formula, but I can't do it. Slowly I am thinking that this formula is wrong. Maybe there is some expert who could tell me if I am right. I have the following problem:

In the end I want to reach the form:

$$ L_{BC} = \frac{L_{AC}\cos{\alpha} - L_{AC'}}{\sin{\alpha}} $$

starting with the formula for similar triangles:

$$ \frac{L_{AC}}{\sin{\theta}} = \frac{L_{AC'}}{\sin{( \theta - \alpha )}} $$

When I combine these two formulas I come to the point that

$$ L_{BC} = L_{AC'} \frac{\cos\theta}{\sin(\theta - \alpha)} $$

Now I don't see any way to replace $ \theta $ so that I am only dependent on the known variables: $$ L_{AC} \hspace{1cm} L_{AC'} \hspace{1cm} \alpha $$

Also expanding the fractions with sin / cos brings me to an deadend. Am I not seeing an obvious connection in these triangles or is there really something wrong about the formula? To make the actual question more clear: I want to calculate $L_{BC}$ using only $\alpha$ , $L_{AC'}$ and $L_{AC}$! And yes, we have $L_{AB}=L_{AB′}$! Thanks!

Answer 1

Let $AB = AB'\equiv x$ then $$ BC = x\cos\theta \\AC = x\sin\theta $$

And $$ AC'=x \sin(\theta-\alpha) \\ \implies AC'=x\sin\theta \cos\alpha -x\cos\theta \sin \alpha \\AC'= AC \cos\alpha - BC\sin\alpha $$

Answer 2

I am going to guess that $AB = AB'$

In which case $AC,AC', BC, BC'$ are proportional to $\cos\theta, \cos (\theta + \alpha),\sin\theta, \sin(\theta + \alpha)$

And, you are trying to show.

$\sin \theta = \frac {cos\theta\cos\alpha - \cos (\theta+\alpha)}{\sin\alpha}$

Which simplifies to $\cos (\theta+\alpha) = cos\theta\cos\alpha - \sin\theta\sin\alpha$

Which is one of your basic trig identities.

and $ABC$ and $AB'C'$ are not similar triangles and $\frac {\cos \theta}{\sin \theta} = \frac{\cos (\theta + \alpha)}{\sin (\theta - \alpha)}$ is incorrect.