Geometry question (sketch included)

[member 428835]

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.

 

[mfb]

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.

 

[member 428835]

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]

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.

 

[member 428835]

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: $R = h\sin\beta/\cos\alpha$.

 

[mfb]

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?

 

[member 428835]

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]

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).

 

[member 428835]

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!