shootforthestars
  • shootforthestars
A company that manufactures custom aluminum windows makes an eyebrow window that is placed on top of a rectangular window. When a customer orders an eyebrow window, the customer gives the width w and height h of the eyebrow. To make the window, the shop needs to know the radius of the circular arc and the length of the circular arc. a) If the width and height of the eyebrow are 36 in. and 10 in., respectively, then what is the radius of the circular arc? b) Find the length of the circular arc for a width of 36 in. and a height of 10 in.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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shootforthestars
  • shootforthestars
c) Find a formula that expresses the radius of the circular arc r in terms of the width of the eyebrow w and the height of the eyebrow h. d) Find a formula that expresses the length of the circular arc L in terms of w and h.
tkhunny
  • tkhunny
Show your work. Express your thoughts. Demonstrate what you can do. Go!
shootforthestars
  • shootforthestars
The answers are supposed to be: a) 21.2 b) 43 c) ((4h^2)+(w^2)/8h) d) ((w^h)/(h^2)+(0.25w^2)) (((h^2)+(25w^2)/h) arcsin (wh)/(h^2)+(0.25w^2))

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shootforthestars
  • shootforthestars
I just don't know how to get them..
anonymous
  • anonymous
I guess eyebrow window is like a segment of the circle, with may or may not be a semi circle, now u can do it
shootforthestars
  • shootforthestars
I know that the window is a rectangle with a semi circle on top, but I still can't figure out how I can find the answer
mathstudent55
  • mathstudent55
Solve for r. |dw:1446352201528:dw|
shootforthestars
  • shootforthestars
Thank you! I found a!
tkhunny
  • tkhunny
You should not be trying to "get them". Most of "them" are wrong. You should be trying to solve the problem and find which is correct. There are a few situations where looking at the possible answers is a good thing to do up front. Very few.
shootforthestars
  • shootforthestars
They are the correct answers
alekos
  • alekos
would you like me to show you how they are done?
shootforthestars
  • shootforthestars
Please! :)
alekos
  • alekos
Lets do b The formula for arc length is s = rθ where s is the arc length and θ is the angle subtended by the arc. We know r so we just have to work out θ
shootforthestars
  • shootforthestars
ok, is the r we're using 21.2?
alekos
  • alekos
Yes r - 21.2 Now the angle at the centre of the circle is θ/2 so sin(θ/2) = 18/r and θ/2 = arcsin(18/r) From here you can find θ in radians
shootforthestars
  • shootforthestars
I got 116.2 for theta
alekos
  • alekos
no thats in degrees. we need radians
shootforthestars
  • shootforthestars
Oh! I found b, so theta is 2.028 and then I multiplied that by 21.2, and got 43! thank you!
alekos
  • alekos
yes, well done. now for c
shootforthestars
  • shootforthestars
ok
alekos
  • alekos
Using pythagoras we have r^2=(r-h)^2+(w/2)^2 => r^2 = r^2-2hr+h^2+(w/2)^2 => 2hr = h^2 + w^2/4 can you see that so far?
shootforthestars
  • shootforthestars
I got h+w^2/8
shootforthestars
  • shootforthestars
I mean h/2 + (w^2)/8h
alekos
  • alekos
yes that's right! and if you put that on a common denominator of 8h you will get your original answer
alekos
  • alekos
happy with that?
shootforthestars
  • shootforthestars
can you show me how I can get my original answer?
alekos
  • alekos
r = h/2 + w^2/8h => r = (4h^2 + w^2)/8h
shootforthestars
  • shootforthestars
oh, I got it now. Thanks!
shootforthestars
  • shootforthestars
Now, I just need help with D
alekos
  • alekos
I got a slightly different expression for D
alekos
  • alekos
Your D answer is wrong Using L = rθ and sinθ/2 = w/2r => θ = 2arcsin(w/2r) => L = 2r.arcsin(w/2r) Now we substitute our previous expression for r Follow so far?
shootforthestars
  • shootforthestars
would it be this? (((8h^2)+(2w^2))/(16h))arcsin (w)/(((8h^2)+(2w^2))/(16h))
alekos
  • alekos
No, we get this........ L = [(4h^2 + w^2)/4h].arcsin[(4hw)/(4h^2 +w^2)]
shootforthestars
  • shootforthestars
Thanks, I got it!
mathstudent55
  • mathstudent55
|dw:1446407245958:dw|
mathstudent55
  • mathstudent55
\(\alpha = \sin^{-1} \dfrac{18}{21.2} \) \(\alpha = 1.014 ~rad\) \(2 \alpha = 2.028 ~rad\) \(s = r \theta\) \(s = r \times 2 \alpha\) \(s = 21.2 \times 2.028\) \(s = 43.0\)
alekos
  • alekos
isnt that a bit late maths student?

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