Pre-Calc Exam Notes 51

Pre-Calc Exam Notes - Solve the triangle △ ABC given a = 5 b = 3 and C = 96 ◦ Solution A B C = 180 ◦ so A B = 180 ◦ − C = 180 ◦ − 96

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The Law of Tangents Section 2.3 51 2.3 The Law of Tangents We have shown how to solve a triangle in all four cases discussed at the beginning of this chapter. An alternative to the Law of Cosines for Case 3 (two sides and the included angle) is the Law of Tangents : Theorem 2.3. Law of Tangents: If a triangle has sides of lengths a , b , and c opposite the angles A , B , and C , respectively, then a b a + b = tan 1 2 ( A B ) tan 1 2 ( A + B ) , (2.17) b c b + c = tan 1 2 ( B C ) tan 1 2 ( B + C ) , (2.18) c a c + a = tan 1 2 ( C A ) tan 1 2 ( C + A ) . (2.19) Note that since tan ( θ ) =− tan θ for any angle θ , we can switch the order of the letters in each of the above formulas. For example, we can rewrite formula (2.17) as b a b + a = tan 1 2 ( B A ) tan 1 2 ( B + A ) , (2.20) and similarly for the other formulas. If a > b , then it is usually more convenient to use formula (2.17), while formula (2.20) is more convenient when b > a . Example 2.10 b = 3 a = 5 c A B C = 96 Case 3: Two sides and the included angle.
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Unformatted text preview: Solve the triangle △ ABC given a = 5, b = 3, and C = 96 ◦ . Solution: A + B + C = 180 ◦ , so A + B = 180 ◦ − C = 180 ◦ − 96 ◦ = 84 ◦ . Thus, by the Law of Tangents, a − b a + b = tan 1 2 ( A − B ) tan 1 2 ( A + B ) ⇒ 5 − 3 5 + 3 = tan 1 2 ( A − B ) tan 1 2 (84 ◦ ) ⇒ tan 1 2 ( A − B ) = 2 8 tan 42 ◦ = 0.2251 ⇒ 1 2 ( A − B ) = 12.7 ◦ ⇒ A − B = 25.4 ◦ . We now have two equations involving A and B , which we can solve by adding the equations: A − B = 25.4 ◦ A + B = 84 ◦ −−−−−−−− 2 A = 109.4 ◦ ⇒ A = 54.7 ◦ ⇒ B = 84 ◦ − 54.7 ◦ ⇒ B = 29.3 ◦ We can ±nd the remaining side c by using the Law of Sines: c = a sin C sin A = 5 sin 96 ◦ sin 54.7 ◦ ⇒ c = 6.09...
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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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