Pre-Calc Exam Notes 98

Pre-Calc Exam Notes 98 - 0.775 rad CBD = 2(0.775) = 1.550...

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98 Chapter 4 Radian Measure §4.3 Example 4.13 The centers of two circles are 7 cm apart, with one circle having a radius of 5 cm and the other a radius of 3 cm. Find the area K of their intersection. Solution: In Figure 4.3.6(a), we see that the intersection of the two circles is the union of the seg- ments formed by the chord CD in each circle. Thus, once we determine the angles CAD and CBD we can calculate the area of each segment and add those areas together to get K . C D 5 4 A B 7 (a) BAC = 1 2 CAD , ABC = 1 2 CBD 5 C 4 7 A B α β (b) Triangle ABC Figure 4.3.6 By symmetry, we see that BAC = 1 2 CAD and ABC = 1 2 CBD . So let α = BAC and β = ABC , as in Figure 4.3.6(b). By the Law of Cosines, we have cos α = 7 2 + 5 2 4 2 2(7)(5) = 0.8286 α = 0.594 rad CAD = 2(0.594) = 1.188 rad cos β = 7 2 + 4 2 5 2 2(7)(4) = 0.7143 β =
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Unformatted text preview: 0.775 rad CBD = 2(0.775) = 1.550 rad Thus, the area K is K = (Area of segment CD in circle at A ) + (Area of segment CD in circle at B ) = 1 2 (5) 2 (1.188 sin 1.188) + 1 2 (4) 2 (1.550 sin 1.550) = 7.656 cm 2 . Exercises For Exercises 1-3, nd the area of the sector for the given angle and radius r . 1. = 2.1 rad, r = 1.2 cm 2. = 3 7 rad, r = 3.5 ft 3. = 78 , r = 6 m 4. The centers of two belt pulleys, with radii of 3 cm and 6 cm, respectively, are 13 cm apart. Find the total area K enclosed by the belt. 5. In Exercise 4 suppose that both belt pulleys have the same radius of 6 cm. Find the total area K enclosed by the belt....
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