College Algebra & Trigonometry, 2018a

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11.4. APPLICATIONS 497 C a =25 135 ◦ b =? B c =50 A We can use the Law of Cosines to solve this problem: b 2 = a 2 + c 2 − 2ac cos B b 2 =25 2 +50 2 − 2 ∗ 25 ∗ 50 ∗ cos 135 ◦ b 2 ≈ 625 + 2500 − 2500 ∗ (−0.7071) b 2 ≈ 3125 + 1767.75 b 2 ≈ 4892.75 b ≈ 69.9 miles
498 CHAPTER 11. THE LAW OF SINES THE LAW OF COSINES Example 2 A pilot flies an airplane in a straight path for 2.5 hours and then makes a course correction, heading 10 ◦ to the left of the original course. The pilot then flies in this direction for 1 hour. If the speed of the plane is a constant 350 mph, how far is the plane from its starting position? Again, we’ll start by making a diagram: C 10 ◦ 170 ◦ B A In this problem, we’ll be working with triangle ABC, shown below. We can calculate the lengths of AB and BC from the information given in the problem and use this to calculate the length of AC: C a = 350 b ∠B = 170 ◦ c = 875 A
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College Algebra & Trigonometry
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c○2018 Richard W. Beveridge This
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4 CONTENTS 2.4 Solution of Rational
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6 CONTENTS 8.2 The Trigonometric Ra
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8 CONTENTS
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College Algebra & Trigonometry, 2018a

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