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.


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.