Partial circumference of a circle

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Hello, I'm sure this is easy for y'all, but I can't figure it out. Can you solve for X for the attached?
 

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Hello, I'm sure this is easy for y'all, but I can't figure it out. Can you solve for X for the attached?
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Locate the center of the circle and join the center and the end-points of the red arc. What is the measure of the central angle.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem

1658104935525.png
 
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Subhotosh Khan said:
Locate the center of the circle and join the center and the end-points of the red arc. What is the measure of the central angle.
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Is there a tool I can use to do this? I don't have access to any.
Subhotosh Khan said:
Please show us what you have tried and exactly where you are stuck.
Click to expand...
I have tried to take 1/3 of the circumference, but that was obviously wrong. I'm exactly stuck at knowing how to approach this problem.
pka said:
You need to read circular segment : [imath]h=8.67 in.[/imath]
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Thank you for this! R = 13in, h = 8.67in, d = 4.33in. Ø, C, and S = Unknown.
 
|||||||E||||||| said:
Is there a tool I can use to do this? I don't have access to any.

I have tried to take 1/3 of the circumference, but that was obviously wrong. I'm exactly stuck at knowing how to approach this problem.

Thank you for this! R = 13in, h = 8.67in, d = 4.33in. Ø, C, and S = Unknown.
Click to expand...
Use this right triangle to find [imath]\theta[/imath]:

1658149035094.png

Then the arc subtends [imath]2\theta[/imath].
 
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Is there a tool I can use to do this? I don't have access to any.
Click to expand...
As shown in the response above (#5), you need to locate the center "approximately" - then apply known trigonometric properties to calculate appropriate unknown variables. You don't need any special tool except pencil/paper.
 
Dr.Peterson said:
Looks good, except for that mistake of calling the arc length a "partial diameter"!

Here is a GeoGebra drawing of it, for confirmation...
Click to expand...

Wow, yeah...I had a typo there. Partial diameter should have been partial circumference/arc length.

Thanks for the confirmation Dr. Peterson! Really helpful.

For context, this solution will make felling trees really easy. Acquire the diameter of the tree, input into this solution, cut a string the length of the arc length, mark the center point, aim the center point where you want the tree to fall, wrap the string around the tree, mark the end points of the arc length, and that's your hinge point.
 
|||||||E||||||| said:
Wow, yeah...I had a typo there. Partial diameter should have been partial circumference/arc length.

Thanks for the confirmation Dr. Peterson! Really helpful.

For context, this solution will make felling trees really easy. Acquire the diameter of the tree, input into this solution, cut a string the length of the arc length, mark the center point, aim the center point where you want the tree to fall, wrap the string around the tree, mark the end points of the arc length, and that's your hinge point.
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Okay.

Or, assuming the idea is that you always want to cut 1/3 of the way into the tree, you could calculate the arc as 0.392 times the circumference. (In the example, that's [imath]\frac{32}{26\pi}[/imath]; in general, it's [imath]\frac{\arccos\left(\frac{1}{3}\right)}{\pi}[/imath].)

As long as the tree is nearly circular ...

Looking around, I see recommendations from 1/5 to 1/3, but nothing like your idea.
 
Two of the best arborists that I know swear by 1/2 and 1/4, so I've always done 1/3. Been working well until this point, but this solution will help with precision. Thank you!
 
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