pre calculus problem
Anna decides to approximate the arc length of a sector of a circle of radius 12 units and central angle 3pi/4. She inscribes N identical isosceles triangles in the sector and attempts to sum N line segments of length AB.
a) Show that S(N)=24N sin 3pi/8N where S(N)=the sum of the N segments each of length AB.
See the diagram below.
Originally Posted by Amy
each of the isosolese triangles will look like the diagram. all we need to do is find the base of the triangle, and then multiply it by N to find the approximation of the arc length, since it is N of these bases will be used to approximate it.
Let the base of each triangle be BC as indicated by the diagram.
since the angle that subtends the arc is 3pi/4, the angle at the vertex of each triangle will be 3pi/4N, since we would divide the angle into N equal peices. if we draw a vertical line that bisects the base, it will also bisect the angle at the vertex since the triangle is isoseles. now we can find the length of the base using the sine ratio, since we obtain two right-triangles when we bisect the isosolese, each with an angle of 1/2 * 3pi/4N = 3pi/8N at the vertex.
recall that the radius of the circle is 12, so the hypotenuse will be 12. the base will be 1/2 BC, which will be the opposite side. so we have:
sin(3pi/8N) = opp/hyp = [(1/2)BC]/12
=> sin(3pi/8N) = BC/24
=> BC = 24sin(3pi/8N)
so the length of each of the bases will be 24sin(3pi/8N), so the length of N bases will be 24Nsin(3pi/8N) as desired
tell me if you don't get my explanation
Thank you Jhevon. I understood. So the actual length can be calculated by taking the limits of S(N) n--> infinity. right?
Here's another approach for finding the length of chord AB.
Let O = the vertex of the isosceles triangle.
. . and AB = the base of the triangle.
The vertex angle is: 3π/4N. .OA = OB = 12.
The Law of Cosines:
. . . . . . . .ABČ .= .OAČ + OBČ - 2(OA)(OB)cosθ
We have: .ABČ .= .12Č + 12Č - 2(12)(12)cos(3π/4N)
. . . . . . . . . . . = .288 - 288·cos(3π/4N)
. . . . . . . . . . . = .288·[1 - cos(3π/4N)]
. . . . . . . . . . . = .288·[2·sinČ(3π/8N)] . ← . Half-angle identity
. . . . . . . . . . . = .576·sinČ(3π/8N)
Therefore: .AB .= .24·sin(3π/8N)
Of course, Jhevon's right-triangle approach is faster and simpler.
. . As usual, he did a great job.
Actually, as usual, i like your solution better. it seems more "rigorous" to me for some reason. your calculations ARE your explanation. with mine, you had to have a lot of background knowledge before coming up with the simple equation. i considered using the cosine rule, but for some reason i imagined it being too messy. boy was i wrong!...or maybe it's because you have the ability to make anything look easy. Good Job!
Originally Posted by Soroban
Originally Posted by Amy