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Unformatted text preview: must be a rectangle. In particular, EF = AC , so EF = 6 r 6. By formula (4.4) we know that h DE = EA · ∠ DAE and h FG = BF · ∠ GBF , where the angles are measured in radians. So thinking of angles in radians (using π rad = 180 ◦ ), we see from Figure 4.2.4 that ∠ DAE = π − ∠ EAC − ∠ BAC = π − π 2 − ∠ BAC = π 2 − ∠ BAC , where sin ∠ BAC = BC AB = 3 15 = 0.2 ⇒ ∠ BAC = 0.201 rad. Thus, ∠ DAE = π 2 − 0.201 = 1.37 rad. So since AE and BF are parallel, we have ∠ ABC = ∠ DAE = 1.37 rad. Thus, ∠ GBF = π − ∠ ABC = π − 1.37 = 1.77 rad. Hence, L = 2 ( h DE + EF + h FG ) = 2 (5 (1.37) + 6 r 6 + 8 (1.77)) = 71.41 cm ....
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This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.
 Fall '11
 Dr.Cheun
 Calculus, Arc Length

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