Circle in n Sectors
Date: 7/8/96 at 4:25:1 From: Wong Cheong Siong Subject: Circle in Sectors Dear Dr. Math, A circle is completely divided into n sectors in such a way that the angles of the sectors are in arithmetic progression. If the smallest of these angles is 8 degrees, and the largest is 52 degrees, calculate n. Best Regards, Wong Cheong Siong
Date: 7/8/96 at 5:53:55 From: Doctor Anthony Subject: Re: Circle in n Sectors The sum of the angles must equal 360 degrees, and we use the formula Sum of n terms = (n/2)(first term + last term) (n/2)(8 + 52) = 360 (n/2)*60 = 360 n/2 = 6 and so n = 12
Date: 7/8/96 at 11:24:12 From: Doctor Brian Subject: Re: Circle in n Sectors Let n = the number of intermediate terms. Let d = the common difference between terms. We have 8 + (8 + d) + (8 + 2d) + ... + (8 + nd) + 52 = 360. We also have that (8 + (n+1)d) = 52, because that would be the next term in the arithmetic progression. For the first equation, subtract the first 8 and the 52 to get that (8 + d) + ... + (8 + nd) = 300. Use the fact that the sum from 1 to n is 1/2 * n * (n+1), and use the fact that there are n 8's to get the equation 8n + 1/2 * n * (n+1) * d = 300. From the second equation you can subtract the 8 to get that (n+1)d = 44. This equation can be substituted verbatim into the other equation to get 8n + 1/2 * n * 44 = 300. This gives 8n + 22n = 300, 30n = 300, n = 10, and substituting back, that d = 4. The n = 10 represents the intermediate steps, so there are a total of 12 arcs....8, 12, 16, 20, ... , 52.
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