\[a_{t}=\dfrac{r^{2}}{\tan\left ( \dfrac{\beta}{2} \right )}- \dfrac{\pi \cdot r^{2}\alpha}{360}\]

Here is an equation I derived in a simple manner for finding the area between two tangents on a circle in terms of the radius if given their angle of intersection, as shown above.

In accordance with this diagram:

We can prove with SSS or SAS that the two triangles are congruent. Here, beta is the angle of intersection and beta is the supplement of alpha. All I am doing is finding the areas of both triangles and subtracting from it the area section of our circle, whatever remains is the area between the curves. Nothing fancy, but it is nice to generalize something, and it might come in handy for someone somewhere.

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TopNewestThis formula would look a little neater if you'd have expressed all angles in radians. – Deeparaj Bhat · 1 year, 2 months ago

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@Deeparaj Bhat Right. So I guess: \(\frac {r^{2}}{tan (\frac{\beta}{2})}- \frac{r^{2}\alpha} {2}\). You are right, it looks a lot nicer. I tried also getting a common denominator, but that looks even worse than using degrees. Thank you, sir. – Drex Beckman · 1 year, 2 months ago

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@Drex Beckman On plugging in \(\beta\), the formula becomes,

\[\frac{r^2}{2}\cdot(\tan{\frac{\alpha}{2}}-\frac{\alpha}{2})\] – Deeparaj Bhat · 1 year, 2 months ago

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@Deeparaj Bhat Can you please tell me how you derived that? Sorry, I'm lacking in maths. Is there an identity you used? Also, am I doing something wrong? I tried this with my question: \(\frac{r^{2}}{2}(tan (\frac {\alpha}{2})-\frac{\alpha}{2})\) where \(\alpha = 0.479965 radians\), and I got an answer of \(\approx 0.00236\), which is exactly half the answer. I think it should be r squared itself, not divided by two, right? I tried this with different r, too. Thanks for your help, it is really appreciated! – Drex Beckman · 1 year, 2 months ago

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I used the following two facts for deriving it:

\[\alpha+\beta=\pi\] \[\tan{(\frac{\pi}{2}-x)}=\cot{x}\]

The first follows by the fact that sum of angles in a quadrilateral ( here ABCD) is \(2\pi\) and since \(\angle ACD=\angle ABD=\frac{\pi}{2}\)

The second is a well known identity which you can find in one of wikis on Brilliant. – Deeparaj Bhat · 1 year, 2 months ago

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– Drex Beckman · 1 year, 2 months ago

Okay, I get it now. I don't really remember lots of trig identities, so I should obviously review them. :P But you did a great job simplifying it. It really does look a lot more beautiful that way. Thanks! :DLog in to reply

– Deeparaj Bhat · 1 year, 2 months ago

In fact, most of the level 5 geometry problems can be solved (almost) exclusively by using trig identities!Log in to reply

– Drex Beckman · 1 year, 2 months ago

Seriously?! Good to know, man. I learned them at school a few years ago, and never used them again. But that was before I thought mathematics was fun. Too bad I didn't get interested in it earlier, because I am way far behind people like you, also I think most of the people on here are smarter than me. xDLog in to reply