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Pre-Calc Exam Notes 60

Pre-Calc Exam Notes 60 - C = 90 ◦ by Thales’ Theorem...

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60 Chapter 2 General Triangles §2.5 To prove this, let O be the center of the circumscribed circle for a triangle ABC . Then O can be either inside, outside, or on the triangle, as in Figure 2.5.2 below. In the first two cases, draw a perpendicular line segment from O to AB at the point D . D O b C a B A c 2 c 2 R R (a) O inside ABC D O b C a B A c 2 c 2 R R (b) O outside ABC O b C a B A c R R (c) O on ABC Figure 2.5.2 Circumscribed circle for ABC The radii OA and OB have the same length R , so AOB is an isosceles triangle. Thus, from elementary geometry we know that OD bisects both the angle AOB and the side AB . So AOD = 1 2 AOB and AD = c 2 . But since the inscribed angle ACB and the central angle AOB intercept the same arc hbracewidest AB , we know from Theorem 2.4 that ACB = 1 2 AOB . Hence, ACB = AOD . So since C = ACB , we have sin C = sin AOD = AD OA = c 2 R = c 2 R 2 R = c sin C , so by the Law of Sines the result follows if O is inside or outside ABC . Now suppose that O is on ABC , say, on the side AB , as in Figure 2.5.2(c). Then AB is a diameter of the circle, so
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Unformatted text preview: C = 90 ◦ by Thales’ Theorem. Hence, sin C = 1, and so 2 R = AB = c = c 1 = c sin C , and the result again follows by the Law of Sines. QED Example 2.17 4 5 3 O A B C Figure 2.5.3 Find the radius R of the circumscribed circle for the triangle △ ABC whose sides are a = 3, b = 4, and c = 5. Solution: We know that △ ABC is a right triangle. So as we see from Figure 2.5.3, sin A = 3/5. Thus, 2 R = a sin A = 3 3 5 = 5 ⇒ R = 2.5 . Note that since R = 2.5, the diameter of the circle is 5, which is the same as AB . Thus, AB must be a diameter of the circle, and so the center O of the circle is the midpoint of AB . Corollary 2.6. For any right triangle, the hypotenuse is a diameter of the circumscribed circle, i.e. the center of the circle is the midpoint of the hypotenuse....
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