Take a sector of this circle with internal angle

[sutupidmath]

Problem: Given a circle of radius 1. Take a sector of this circle with internal angle A, where 0=<A=<pi/2. Find a formula for the radius of the smallest circle that will perfectly fit this sector, as a function of A.

Solution.

I used laws of sine and cosine and came up with:

R=\frac{\sqrt{2-2cos(A)}}{2sin(A)}.

So, i was wondering whether this is correct.

Basically, i thought this problem was equivalent to finding the radius of a circle circumscribing a given triangle. But i am not sure whether this is the right approach.

Thanks!

P.S. This is NOT a homework problem.

 

[tiny-tim]

hi sutupidmath!

that's sec(a/2)/2 … that doesn't look right

how did you get this?

 

[zgozvrm]

First of all, what do you mean by "internal angle?"

Since you refer to a sector of a circle, I'm guessing that \angle A[/tex] is a &quot;central angle,&quot; correct?

 

[sutupidmath]

hi sutupidmath!

that's sec(a/2)/2 … that doesn't look right

how did you get this?

Let the sides of the triangle be 1,1 and c, and the angle oposite to c, be A. Then using the law of cosine, we have for c:

c^2=2-2cos(A)

Second, using the laws of sine, we have: \frac{c}{sin(A)}=2R where R is the radius of the outer circle.

Combining these two, i arrived at my R.

 

[sutupidmath]

First of all, what do you mean by "internal angle?"

Since you refer to a sector of a circle, I'm guessing that \angle A[/tex] is a &quot;central angle,&quot; correct?

<br /> <br /> Yes, it is a central angle.

 

[tiny-tim]

… using the law of cosine …

a lot simpler is to bisect A, immediately giving c = 2sin(A/2)

Second, using the laws of sine …

i don't understand this at all …

your LHS is 2sin(A/2)/sinA, which is 1/cos(A/2) …

what does that have to do with R?

and anyway, you're asked to find the inscribed circle

 

[sutupidmath]

a lot simpler is to bisect A, immediately giving c = 2sin(A/2)


i don't understand this at all …

your LHS is 2sin(A/2)/sinA, which is 1/cos(A/2) …

what does that have to do with R?

and anyway, you're asked to find the inscribed circle

I just realized that my answer is the same as yours, up to a few trig identities.

By the law of sine i meant a/sinA=b/sinB=c/sinC=2R, where R is the radius of the circumscribed circle around the triangle ABC.

And, i think we are supposed to find the radius of the circumscribed circle, since we want to fit the sector taken out of the circle with radius 1 in this other circle.

 

[zgozvrm]

Here's a pic of what you described.

Note that the small circle is inscribed within a triangle; 2 sides that intersect at the center of the large circle. The 3rd side is tangent to the point where the 2 circles intersect and, thus, perpendicular to the radius of the large circle at that point.

 

[tiny-tim]

And, i think we are supposed to find the radius of the circumscribed circle, since we want to fit the sector taken out of the circle with radius 1 in this other circle.

but the circumscribed circle is the original circle of radius 1

and "fit" usually means inside …

Problem: Given a circle of radius 1. Take a sector of this circle with internal angle A, where 0=<A=<pi/2. Find a formula for the radius of the smallest circle that will perfectly fit this sector, as a function of A.

 

[zgozvrm]

but the circumscribed circle is the original circle of radius 1

Actually the circumscribed circle (the circle that circumscribes the described sector is smaller than the original circle (see attachment).

 

[zgozvrm]

"fit" usually means inside …

The circle in my last post "fits" around the sector.

Since the OP stated "the smallest circle that will perfectly fit this sector," I believe that is the circle he is looking for (not the first picture I posted).

 

[tiny-tim]

oh yes, of course!

 

[zgozvrm]

"fit" usually means inside …

... and, in this case, we're talking about how the sector perfectly fits inside the circle.

 

[sutupidmath]

Well, my answer was correct! :)

Thanks for your input guys.

 

[zgozvrm]

Well, my answer was correct! :)

Yep!

 

[zgozvrm]

There are more "compact" answers...
I came up with

R = \frac{\sin (A/2)}{\sin A}