Geometry question (sketch included)

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In summary, the conversation discussed the use of law of sines to determine the radius of a circle given certain parameters. The formula used was R = hsinβ/cosα, but the speaker encountered problems when plotting the results. It was suggested to introduce a new angle, α' = α - 90°, to better fit with the sketch. Ultimately, the speaker realized that using arccos instead of arctan for calculating the values of θ between two walls resulted in more accurate results.
  • #1
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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.
 

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  • #2
I moved the thread to our homework section.
joshmccraney said:
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.
 
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  • #3
mfb said:
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.
 
  • #4
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.
joshmccraney said:
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.
 
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  • #5
mfb said:
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##?

mfb said:
##\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##.
 
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  • #6
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?
 
  • #7
mfb said:
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.
 
  • #8
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).
 
  • #9
mfb said:
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!
 

FAQ: Geometry question (sketch included)

1. What is the definition of Geometry?

Geometry is a branch of mathematics that deals with the study of shapes, sizes, and positions of figures in space.

2. What are the different types of angles in Geometry?

There are four types of angles in Geometry: acute angles (less than 90 degrees), obtuse angles (greater than 90 degrees), right angles (exactly 90 degrees), and straight angles (exactly 180 degrees).

3. How do you calculate the area of a triangle?

The formula for calculating the area of a triangle is: (base x height)/2. In other words, you multiply the base of the triangle by its height and then divide the result by 2.

4. What is the Pythagorean Theorem?

The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

5. How does Geometry relate to other fields of study?

Geometry has applications in various fields such as physics, engineering, architecture, and computer graphics. It also plays a crucial role in understanding and analyzing patterns and relationships in nature and the universe.

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