Pre-Calc Exam Notes 44

Pre-Calc Exam Notes 44 - If a triangle has sides of lengths...

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44 Chapter 2 General Triangles §2.2 2.2 The Law of Cosines We will now discuss how to solve a triangle in Case 3: two sides and the angle between them. First, let us see what happens when we try to use the Law of Sines for this case. Example 2.4 b = 4 a c = 5 A = 30 B C Case 3: Two sides and the angle between them. Solve the triangle ABC given A = 30 , b = 4, and c = 5. Solution: Using the Law of Sines, we have a sin 30 = 4 sin B = 5 sin C , where each of the equations has two unknown parts, making the problem impossible to solve. For example, to solve for a we could use the equation 4 sin B = 5 sin C to solve for sin B in terms of sin C and substitute that into the equation a sin 30 = 4 sin B . But that would just result in the equation a sin 30 = 5 sin C , which we already knew and which still has two unknowns! Thus, this problem can not be solved using the Law of Sines. To solve the triangle in the above example, we can use the Law of Cosines : Theorem 2.2. Law of Cosines:
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Unformatted text preview: If a triangle has sides of lengths a , b , and c opposite the angles A , B , and C , respectively, then a 2 = b 2 + c 2 − 2 bc cos A , (2.9) b 2 = c 2 + a 2 − 2 ca cos B , (2.10) c 2 = a 2 + b 2 − 2 ab cos C . (2.11) To prove the Law of Cosines, let △ ABC be an oblique triangle. Then △ ABC can be acute, as in Figure 2.2.1(a), or it can be obtuse, as in Figure 2.2.1(b). In each case, draw the altitude from the vertex at C to the side AB . In Figure 2.2.1(a) the altitude divides AB into two line segments with lengths x and c − x , while in Figure 2.2.1(b) the altitude extends the side AB by a distance x . Let h be the height of the altitude. h b a c A B C x c − x (a) Acute triangle h b a c A B C 180 ◦ − B x (b) Obtuse triangle Figure 2.2.1 Proof of the Law of Cosines for an oblique triangle △ ABC...
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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.

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