558 CHAPTER 5 Trigonometric Functions (b) Use the identity cos u = 1 sec u . If sec u = 1.0545829, then cos u = 1 1.0545829 . Now, find u using the inverse cosine function. u = cos-1 a 1 1.0545829b ≈18.514704° See Figure 40 on the previous page. S Now Try Exercises 125 and 129. CAUTION Compare Examples 8(b) and 9(b). • To determine the secant of an angle, as in Example 8(b), we find the reciprocal of the cosine of the angle. • To determine an angle with a given secant value, as in Example 9(b), we find the inverse cosine of the reciprocal of the value. An Application EXAMPLE 10 Finding Grade Resistance When an automobile travels uphill or downhill on a highway, it experiences a force due to gravity. This force F in pounds is the grade resistance and is modeled by F = W sin u, where u is the grade and W is the weight of the automobile. If the automobile is moving uphill, then u 70°; if downhill, then u 60°. See Figure 41. (Data from Mannering, F., and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.) (a) Calculate F to the nearest 10 lb for a 2500-lb car traveling an uphill grade with u = 2.5°. (b) Calculate F to the nearest 10 lb for a 5000-lb truck traveling a downhill grade with u = -6.1°. (c) Calculate F for u = 0° and u = 90°. Do these answers agree with intuition? SOLUTION (a) F = W sin u Given model for grade resistance F = 2500 sin 2.5° Substitute given values. F ≈110 lb Evaluate. (b) F = W sin u = 5000 sin1-6.1°2 ≈ -530 lb F is negative because the truck is moving downhill. (c) F = W sin u = W sin 0° = W102 = 0 lb F = W sin u = W sin 90° = W112 = W lb This agrees with intuition because if u = 0°, then there is level ground and gravity does not cause the vehicle to roll. If u were 90°, the road would be vertical and the full weight of the vehicle would be pulled downward by gravity, so F = W. S Now Try Exercises 135 and 137. u > 0° u < 0° Figure 41
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