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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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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, Angles

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