# Geometry question (sketch included)

Gold Member
Attached is a sketch of geometry in question, sketch.pdf. I am trying to determine the circle's radius ##R## given half corner angle ##\beta##, contact angle ##\alpha##, and displaced vertex ##h##. What I find from law of sines is ##R = h \sin\beta / \cos\alpha##.

However, when I plot this I do not always get a desired answer. See case.pdf. Here I set ##\alpha=60^\circ## yet obviously in this case ##\alpha > 90^\circ##. In this case ##\alpha=120^\circ##.

Any help here is greatly appreciated. I know law of sines has 0,1,2 cases, though I can't see how that is applicable here.

Edit: For what it's worth, it is clear in the limit case ##h\to0\implies\alpha\to 90^\circ##. Just an observation.

#### Attachments

• sketch.pdf
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• case.pdf
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Last edited:

## Answers and Replies

mfb
Mentor
I moved the thread to our homework section.
Here I set ##\alpha=60^\circ##
Why would you do that? Just use the correct angle. Or introduce ##\alpha' = \alpha - 90^\circ## which is the angle in your triangle.

scottdave
Gold Member
Why would you do that? Just use the correct angle. Or introduce ##\alpha' = \alpha - 90^\circ## which is the angle in your triangle.
Can you elaborate, I'm very confused what you mean by "correct angle".

My idea is to choose any ##\alpha##, so why couldn't I choose ##\alpha = 60^\circ##?

Edit: I see I may have been confusing when I spoke about ##\alpha = 60^\circ##. What I mean is when i set ##\alpha=60## the graphical result measures ##\alpha## to be ##120^\circ##, so there is an problem.

mfb
Mentor
You have ##\alpha## and you have the interior angle in the triangle, these are different angles, and it is probably advisable to introduce a name for the latter.
What I mean is when i set ##\alpha=60## the graphical result measures ##\alpha## to be ##120^\circ##, so there is an problem.
##\alpha < 90^\circ## doesn't fit to your sketch and you will be at the left side of the circle.

joshmccraney
Gold Member
You have ##\alpha## and you have the interior angle in the triangle, these are different angles, and it is probably advisable to introduce a name for the latter..
Which triangle are you referring to? Actually rereading your earlier post about ##\alpha'## it is obvious what you're referring to. Does it really require a name since we are taking ##\sin(\alpha - 90^\circ) = \cos\alpha##?

##\alpha < 90^\circ## doesn't fit to your sketch and you will be at the left side of the circle.
Right, ##\alpha < 90^\circ## would be the left side. However, the conclusion still holds: ##R = h\sin\beta/\cos\alpha##.

scottdave
mfb
Mentor
I'm confused. Let's start from scratch:
What exactly are the parameters you have trouble with? What do you set for ##\beta## and ##\alpha##, what do you expect for R/h, what do you get?

Gold Member
I'm confused. Let's start from scratch:
What exactly are the parameters you have trouble with? What do you set for ##\beta## and ##\alpha##, what do you expect for R/h, what do you get?
The geometric problem is as stated above. What I would like to do is choose any ##\alpha,\beta,h## and from these three parameters determine a radius ##R##. I thought this was ##R = h\sin\beta/\cos\alpha##.

After this I would like to find the values of ##\theta## (polar coordinates) so that the circle lies just between the corner edges, not overlapping as shown above. If this makes sense I'll continue with where I am at, as I think I have the ##\theta## values determined, though there are 3 different cases, which is kind of messy.

mfb
Mentor
What is a set of ##\beta##, ##\alpha## where you ran into a problem with that formula? You only gave an ##\alpha## value (actually, two at the same time).

Gold Member
What is a set of ##\beta##, ##\alpha## where you ran into a problem with that formula? You only gave an ##\alpha## value (actually, two at the same time).
Sorry, I think I figured everything out. See, I was initially confused because when calculating the values for ##\theta## to go from one wall (straight line) to another I was using the arctangent function. However, after thinking about the outputs of arctan i switched to arccos and everything came out very well. I can elaborate further but I think I am good now.

Thanks for your help, and thanks for pointing out the two triangles!