812 CHAPTER 8 Applications of Trigonometry Horizontal and Vertical Components The horizontal and vertical components, respectively, of a vector u having magnitude ∣ u∣ and direction angle u are given by the following. a = ∣ u∣ cos U and b = ∣ u∣ sin U That is, u = 8a, b9 = 8 u cos u, u sin u9. EXAMPLE 2 Finding Horizontal and Vertical Components Vector w in Figure 36 has magnitude 25.0 and direction angle 41.7°. Find the horizontal and vertical components. ALGEBRAIC SOLUTION Use the formulas below, with w = 25.0 and u = 41.7°. a = w cos u a = 25.0 cos 41.7° a ≈18.7 b = w sin u b = 25.0 sin 41.7° b ≈16.6 Therefore, w= 818.7, 16.69. The horizontal component is 18.7, and the vertical component is 16.6 (rounded to the nearest tenth). GRAPHING CALCULATOR SOLUTION See Figure 37. The results support the algebraic solution. Figure 37 S Now Try Exercise 15. y x 0 25.0 b a w 41.7° Figure 36 EXAMPLE 3 Writing Vectors in the Form 8 a, b 9 Write each vector in Figure 38 in the form 8a, b9. SOLUTION u = 85 cos 60°, 5 sin 60°9 = h5 # 1 2 , 5 # 23 2 i = h 5 2 , 523 2 i v = 82 cos 180°, 2 sin 180°9 = 821-12, 21029 = 8-2, 09 w= 86 cos 280°, 6 sin 280°9 ≈ 81.0419, -5.90889 Use a calculator. S Now Try Exercises 21 and 23. y x 0 u v w 2 5 6 280° 60° 180° Figure 38 Operations with Vectors As shown in Figure 39, m= 8a, b9, n = 8c, d9, and p = 8a + c, b + d9. Using geometry, we can show that the endpoints of the three vectors and the origin form a parallelogram. A diagonal of this parallelogram gives the resultant of m and n, so we have p = m+ n or 8a + c, b + d9 = 8a, b9 + 8c, d9. Similarly, we can verify the following operations. x 0 ka, bl m p n kc, dl ka + c, b + dl y Figure 39
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