2
$\begingroup$

So I am dealing with the hexagon as shown in the picture below and I need to find out how one angle depends on another angle. More specifically, I need $\frac{d\psi}{d\varphi}$ at $\varphi=0$.

enter image description here

Note that except for $\psi$ and $\varphi$, all angles are fixed and that $K$ and $L$ are fixed lengths. The vertex at the green angle $\gamma$ is uniquely defined since we are given four lengths and all six angles of a hexagon. Note that this is true since I know (from context) that the angles are chosen so that the green angle $\gamma$ is not equal to $\pi$ (which would mean that the hexagon is actually a pentagon in which case that point would of course not be unique).

Clearly if $\varphi=0$, then the picture is symmetric and $\psi=\gamma/2$. That one is easy. What I need, though, is the rate of change $\frac{d\psi}{d\varphi}$ at $\varphi=0$.

I found a solution but it is terribly ugly, so I was wondering if I ignored some basic trigonometry trick that simplifies this issue.

Here is my solution. Consider this picture with some added info:

enter image description here

  • By the law of cosines, $c^2=L^2+K^2-2KL\cos(\pi/6+\varphi)$, so we get $c$.
  • By the law of cosines, $d^2=L^2+K^2-2KL\cos(\pi/6-\varphi)$, so we get $d$.
  • By the law of sines, $\frac{\sin(\delta)}{K}=\frac{\sin(\varphi+\pi/6)}{c}$, so we get $\delta$.
  • By the law of sines, $\frac{\sin(\epsilon)}{K}=\frac{\sin(\varphi-\pi/6)}{d}$, so we get $\epsilon$.

Now we can combine the following two equations from the law of sines:

  • $\frac{\sin \psi}{c}=\frac{\sin(\alpha-\delta)}{\ell}$
  • $\frac{\sin (\gamma-\psi)}{d}=\frac{\sin(\alpha-\epsilon)}{\ell}$

Combining them by eliminating $\ell$, we get $\frac{\sin(\alpha-\delta)}{\sin\psi}c=\frac{\sin(\alpha-\epsilon)}{\sin(\gamma-\psi)}d$.

Now $\psi$ is the only unknown and I can implicitly differentiate the hell out of these equations and get some unwieldly formula for $d\psi/d\phi$ at $\phi=0$.

I had to deal with a similar problem a while ago and at first I had a horrible formula only to later realize if only I had used the law of sines differently in one step, the equation would be much simpler.

So I am wondering if there is a better way to do this besides the one above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.