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August 5th, 2019, 07:09 AM   #1
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Problem involving circumferences and trapazoids

Hi my name is Ed and have a problem involving circumferences and trapazoids which I need some help in solving. In the diagram I've including are two trapazoids with circles inside of them. The distance from a to b is equal to 33.8 meters, from b to c is equal to 23.96 meters and from c to d is equal to 34.4 meters.

Given those measures, how does one determine ...
1. The angle of ray af
2. The diameter of the blue circle
3. The sizes of the two trapazoids the green and blue circles are contained within.

Kind regards
Ed
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Last edited by edwin1576; August 5th, 2019 at 07:11 AM. Reason: corrected distance from a to be .... to a to b
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August 5th, 2019, 09:20 AM   #2
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Quote:
The distance from a to b is equal to 33.8 meters, from b to c is equal to 23.96 meters and from c to d is equal to 34.4 meters.

Given those measures, how does one determine ...
1. The angle of ray af
2. The diameter of the blue circle
3. The sizes of the two trapazoids the green and blue circles are contained within.


1. Ray af makes an angle $\theta = \arcsin\left(\dfrac{r}{r+33.8}\right)$, where $r$ is the small circle radius.

2. blue circle radius (and hence, diameter) can be found using a sine ratio and $\theta$ in a similar manner.

3. What do you mean by size of the trapezoids? ... area? side lengths? something else?

btw ... this problem should have been posted in the trigonometry section. I'm sure one of the mods will move it there eventually.
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August 5th, 2019, 09:50 AM   #3
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Thank you for the reply Skeeter.

Could you please expand on (2. blue circle radius (and hence, diameter) can be found using a sine ratio and θ in a similar manner.)

As far as the size of trapezoids are concerned, I am looking for the length of their sides which form the perimeter around the circles. How does one go about calculating that ?

Kind regards
Ed

Last edited by edwin1576; August 5th, 2019 at 10:04 AM. Reason: needed to post additional query
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August 5th, 2019, 10:24 AM   #4
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Originally Posted by edwin1576 View Post
Could you please expand on (2. blue circle radius (and hence, diameter) can be found using a sine ratio and θ in a similar manner.)
Once you determine the value of $\theta$, you can determine the value of the large circle radius, $\color{red}R$, (and hence, its diameter) as shown in the attached ...
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August 5th, 2019, 11:05 AM   #5
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Hi Skeeter.

Does not Sin θ = R/(R+ab) imply that one already knows what the value of R is ? I mean in essence all you have is the value of ab and the angle of the ray af. Could you please show my how get to the value of R.

Kind regards
Ed
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August 5th, 2019, 11:28 AM   #6
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Originally Posted by edwin1576 View Post
Does not Sin θ = R/(R+ab) imply that one already knows what the value of R is ?
it's $\sin{\theta} = \dfrac{R}{R+ad}$ ... ad, not ab.

You determine $\theta$ from step 1. You know ad = ab + bc + cd.

Do the algebra and solve for $R$, the radius of the large circle.
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August 5th, 2019, 12:04 PM   #7
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Hi Skeeter.

My bad, ad instead of ab. My problem is the algebra, could you please assist me with that ?
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August 5th, 2019, 12:57 PM   #8
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$\sin{\theta} = \dfrac{R}{R+ad}$

multiply both sides by $(R+ad)$ ...

$\sin{\theta}(R+ad) = \dfrac{R}{R+ad} \cdot (R+ad)$

$\sin{\theta}(R+ad) = \dfrac{R}{\cancel{R+ad}} \cdot (\cancel{R+ad})$

distribute the left side ...

$R\sin{\theta} + ad \cdot \sin{\theta} = R$

subtract $R\sin{\theta}$ from both sides ...

$ad \cdot \sin{\theta} = R - R\sin{\theta}$

factor out the common factor $R$ from the two terms on the right side ...

$ad \cdot \sin{\theta} = R(1-\sin{\theta})$

divide both sides by $(1-\sin{\theta})$ ...

$\dfrac{ad \cdot \sin{\theta}}{1-\sin{\theta}} = R$
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August 5th, 2019, 01:07 PM   #9
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Hi Skeeter. Thank you so much for help. I've got it now.

Kind Regards
Ed
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