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Pre-Calc Exam Notes 40

# Pre-Calc Exam Notes 40 - 40 Chapter 2 General Triangles 2.1...

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40 Chapter 2 General Triangles §2.1 Example2.1 b a = 10 c A = 41 B C = 75 Case 1: One side and two angles. Solve the triangle ABC given a = 10, A = 41 , and C = 75 . Solution: We can find the third angle by subtracting the other two angles from 180 , then use the law of sines to find the two unknown sides. In this example we need to find B , b , and c . First, we see that B = 180 A C = 180 41 75 B = 64 . So by the Law of Sines we have b sin B = a sin A b = a sin B sin A = 10 sin 64 sin 41 b = 13 . 7 , and c sin C = a sin A c = a sin C sin A = 10 sin 75 sin 41 c = 14 . 7 . Example2.2 b = 30 a = 18 c A = 25 B C Case 2: Two sides and one opposite angle. Solve the triangle ABC given a = 18, A = 25 , and b = 30. Solution: In this example we know the side a and its opposite angle A , and we know the side b . We can use the Law of Sines to find the other opposite angle B , then find the third angle C by subtracting A and B from 180 , then use the law of sines to find the third side c . By the Law of Sines, we have sin B b = sin A a sin B = b sin A a = 30 sin 25 18 sin B = 0 . 7044 . Using the a0 sin 1 button on a calculator gives B = 44 . 8 . However, recall from Section 1.5 that sin (180 B ) = sin B . So there is a second possible solution for
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