I apologise for the lack of a precise term or title, math isn't my strong suit.

I'm trying to calculate the length of L, given angle A and radius of circle D, so that lines b and c tangents with circle D.

Image of problem

Can anyone help me?


Hint: Using the definition of sine and the orthogonality of the tangent one has $$\sin\left(\alpha/2\right)=\frac{\text{radius}}{\text{distance from center}}$$

  • 1
    $\begingroup$ It's actually $3/\sin(\dots)$. Moreover, fortunately, since $$\sin(45/2)=\frac{\sqrt{2-\sqrt{2}}}{2},$$ you even have a closed form for the distance $\endgroup$ – b00n heT Jul 31 '19 at 13:18
  • $\begingroup$ Sorry I messed up my previous comment. To repeat: So given radius 3 and angle 45 that would be $$\sin\left(45/2\right)=\frac{3}{\text{distance from center}}$$ And that isolated would be $$\text{distance from center}=\frac{3}{\sin\left(45/2\right)}$$ $\endgroup$ – Martin Jul 31 '19 at 13:22
  • $\begingroup$ That's correct. $\endgroup$ – b00n heT Jul 31 '19 at 13:24
  • $\begingroup$ Thank you very much. Sorry, first time formatting equations, so that took at bit. No previews for comments. :) $\endgroup$ – Martin Jul 31 '19 at 13:24
  • $\begingroup$ No worries! glad to be of help $\endgroup$ – b00n heT Jul 31 '19 at 13:24

Supplementary to b00n heT's answer:

Let's call the intersection of $b$ with the circunference $R$ and $r=d(R,D)$, the radius of the circumference.

Considering the right-angle triangle [ARD], we have that $\sin(RÂD)=\frac{r}{L}$, from the definition of the $\sin$ of an angle.

Then, if $A=2RÂD$, then: $$\sin\left(\frac{A}{2}\right)=\frac{r}{L} \Leftrightarrow L=\frac{r}{\sin\left(\frac{A}{2}\right)}$$

  • $\begingroup$ But in my example [ARD] isn't a right angled triangle, right? $\endgroup$ – Martin Aug 1 '19 at 11:48
  • $\begingroup$ $b=AR$, $b$ is tangent to the circumference in the point $R$, therefore, $RD\perp b$, so, $\sphericalangle ARD$ is a right angle, its amplitude is 90 degrees. Note that the point $R$ is not represented in your picture, I "created" that point. So, $[ARD]$ is a right angled triangle. See tangent line in Wikipedia for more information. $\endgroup$ – t3m2 Aug 1 '19 at 12:07
  • 1
    $\begingroup$ Ah yeah, makes sense. I don't know what I was imagining. It's obviously right angled. :D $\endgroup$ – Martin Aug 1 '19 at 12:11

L = R / Sin(A/2)

where R is the radius of the circle and A is the angle


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