$Calculate the angle of an isosceles triangle to cover a distance on a plane$

Question

$Calculate the angle of an isosceles triangle to cover a distance on a plane$

Knowing the points a, b and the angle alpha, calculate the necessary angle beta to cover a known distance C, on the Y plane.

The ab line segment is part of the triangle bisector.

Trigonometry

Luis Fonseca

Matchmaticians Calculate the angle of an isosceles triangle to cover a distance on a plane File #1

File #1 (png)

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Erdos

0

There doesn't seem to be enough information to solve this. Is the line from a to b is the bisector of the triangle? What do we know about the line from a to b?
- Luis Fonseca
  
  0
  
  That's correct. A to B is the bisector of the triangle and we know its length since we know both points.

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Erdos

0

There doesn't seem to be enough information to solve this. Is the line from a to b is the bisector of the triangle? What do we know about the line from a to b?

Luis Fonseca

0

That's correct. A to B is the bisector of the triangle and we know its length since we know both points.
Luis Fonseca

0

That's correct. A to B is the bisector of the triangle and we know its length since we know both points.

Answer 1

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I refer to the attached picture for lines $L_1$ and $L_2$.

The slope of the line $L_1$ is $\tan(\alpha-\frac{\beta}{2})$ and this line passes through $a=(a_1, a_2)$. Hence the equation of the line is
\[y=\tan(\alpha-\frac{\beta}{2}) (x-a_1)+a_2.\]
Similarly the equation of the line $L_2$ is given by
\[y=\tan(\alpha+\frac{\beta}{2}) (x-a_1)+a_2.\]

Then the distance $c$ is just the difference of the $y$ values of these lines at $x=b_1$, where $b=(b_1,b_2)$.

Hence
\[c=\tan(\alpha+\frac{\beta}{2}) (b_1-a_1)+a_2 - [\tan(\alpha-\frac{\beta}{2}) (b_1-a_1)+a_2]\]
\[=(b_1-a_1)[\tan(\alpha+\frac{\beta}{2})-\tan(\alpha-\frac{\beta}{2})]\]
\[=(b_1-a_1)[\frac{\tan \alpha +\tan (\frac{\beta}{2})}{1-\tan (\alpha)\tan(\frac{\beta}{2})}-\frac{\tan \alpha -\tan (\frac{\beta}{2})}{1+\tan (\alpha)\tan(\frac{\beta}{2})}]\]
\[=(b_1-a_1)[\frac{\tan \alpha +\tan (\frac{\beta}{2})+\tan^2 (\alpha)\tan(\frac{\beta}{2})+\tan (\alpha)\tan^2(\frac{\beta}{2})}{1-\tan^2 (\alpha)\tan^2(\frac{\beta}{2})}\]
\[-\frac{\tan \alpha -\tan (\frac{\beta}{2})-\tan^2 (\alpha)\tan(\frac{\beta}{2})+\tan (\alpha)\tan^2(\frac{\beta}{2})}{1-\tan^2 (\alpha)\tan^2(\frac{\beta}{2})}]\]
\[=(b_1-a_1)[\frac{2\tan (\frac{\beta}{2})+2\tan^2 (\alpha)\tan(\frac{\beta}{2})}{1-\tan^2 (\alpha)\tan^2(\frac{\beta}{2})}]=c.\]
Hence
\[2\tan (\frac{\beta}{2})(1+\tan^2 (\alpha))=\frac{c}{b_1-a_1}[1-\tan^2 (\alpha)\tan^2(\frac{\beta}{2})]\]
which simplifies to
\[(\frac{c}{b_1-a_1}\tan^2 (\alpha) ) \tan^2(\frac{\beta}{2}) +2\tan (\frac{\beta}{2})(1+\tan^2 (\alpha))-\frac{c}{b_1-a_1}=0, \]
or
\[\tan^2 (\alpha) \tan^2(\frac{\beta}{2}) +2\frac{b_1-a_1}{c}(1+\tan^2 (\alpha))\tan (\frac{\beta}{2})-1=0. (*)\]
Using quadratioc formula we get
\[\tan(\frac{\beta}{2})=\frac{-B\pm\sqrt{B^2-4AC}}{2A},\]
where
\[A=\tan^2 (\alpha), B=2\frac{b_1-a_1}{c}(1+\tan^2 (\alpha)), C=-1.\]
We should use the plus sign in the quadratic formula as the negative sign will lead to a non-admissable solution.

Then
$$\frac{\beta}{2}=\tan^{-1}(\frac{-B\pm\sqrt{B^2-4AC}}{2A}).$$
Multiplying by 2, you will get a formula for $\beta$
$$\beta=2\tan^{-1}(\frac{-B\pm\sqrt{B^2-4AC}}{2A}). (**)$$

************
For $\alpha =0$ the equation (*) becomes
\[2\frac{b_1-a_1}{c}(1+0)\tan (\frac{\beta}{2})-1=0 \]
\[\rightarrow \tan (\frac{\beta}{2})= \frac{c}{2(b_1-a_1)}\]
\[\rightarrow \beta =2 \tan^{-1}(\frac{c}{2(b_1-a_1)}).\]

For $\alpha$ close to $0$ in the formula (**) we are dividing by small numbers, so the computations should be done with high level of accuracy.

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Erdos

Erdos

0

This was a very tricky problem! Let me know if you have any questions.
Luis Fonseca

0

Hi Erdos, thanks for your work. Excuse me for my lack of knowledge here, but shouldn't it be solved for the β beta angle? Also, I'm going to be using this in a game engine (Unity) where I can't seem to find a method for the squared tangent like it shows on your equation. Is it the same as tan(a) * tan(a) ?
- Erdos
  
  +1
  
  I added some more details at the end to show how you get beta. Yes, tan^2 (a)=tan (a) * tan (a).
Luis Fonseca

0

Erdos, at the end you say "multiplying by 2, you will get a formula for β", but I don't see that reflected in the formula. Can you please double-check?
- Erdos
  
  0
  
  I just fixed it.
Luis Fonseca

0

Erdos, can you please tell me if there's a website where I can put this equation and then get a result by passing values to the variables (α, a1, b1 and c)? This would allow me to easily confirm if this returns correct values without having to implement it in the game engine.
- Erdos
  
  +1
  
  This can be done using a calculator: https://www.desmos.com/scientific
Luis Fonseca

0

Just implemented this and it works great, thank you very much!
- Erdos
  
  0
  
  I am glad I was able to help.
Luis Fonseca

0

Erdos, I just realised that for very small numbers of α (between -1 an 1) I get slightly incorrect results. For 0 it returns NaN (not a number). Do you have any clue why this happens?
- Erdos
  
  +1
  
  I added some notes at the end of my solution.

$Calculate the angle of an isosceles triangle to cover a distance on a plane$

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