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.
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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.
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The answers are supposed to be:
d) ((w^h)/(h^2)+(0.25w^2)) (((h^2)+(25w^2)/h) arcsin (wh)/(h^2)+(0.25w^2))
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I just don't know how to get them..
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
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
Solve for r.
Thank you! I found a!
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.
They are the correct answers
would you like me to show you how they are done?
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
ok, is the r we're using 21.2?
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
I got 116.2 for theta
no thats in degrees. we need radians
Oh! I found b, so theta is 2.028 and then I multiplied that by 21.2, and got 43! thank you!
yes, well done.
now for c
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?
I got h+w^2/8
I mean h/2 + (w^2)/8h
yes that's right! and if you put that on a common denominator of 8h you will get your original answer
happy with that?
can you show me how I can get my original answer?
r = h/2 + w^2/8h
=> r = (4h^2 + w^2)/8h
oh, I got it now. Thanks!
Now, I just need help with D
I got a slightly different expression for D
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?
would it be this?
No, we get this........
L = [(4h^2 + w^2)/4h].arcsin[(4hw)/(4h^2 +w^2)]