College Algebra & Trigonometry, 2018a

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11.2. THE LAW OF SINES: THE AMBIGUOUS CASE 479 Two Triangles h
480 CHAPTER 11. THE LAW OF SINES THE LAW OF COSINES Example 1 Solve the triangle if: ∠A = 112 ◦ , a =45, b =24 Round the angles and side lengths to the nearest 10 th . Using the Law of Sines, we can say that: sin 112 ◦ 45 = sin B 24 0.9272 45 ≈ sin B 24 24∗ 0.9272 45 ≈ sin B 0.4945 ≈ sin B Then, we find sin −1 (0.4945) ≈ 29.6 ◦ . Remember from Chapter 3 that there is a Quadrant II angle that has sin θ ≈ 0.4945, with a reference angle of 29.6 ◦ . So, ∠B could also be ≈ 150.4 ◦ . However, with ∠A = 112 ◦ , there is no way that another angle of 150.4 ◦ would fit inside the same triangle. For this reason, we know then that ∠B must be 29.6 ◦ . 29.6 ◦ ≈ B So now ∠A = 112 ◦ ∠B ≈ 29.6 ◦ and ∠C = 180 ◦ − (112 ◦ +29.6 ◦ ) = 180 ◦ − 141.6 ◦ ≈ 38.4 ◦ ∠C ≈ 38.4 ◦ We already know that a =45and b =24. To find side c, I would recommend using the most exact values possible in the Law of Sines calculation. This will provide the most accurate result in finding the length of side c.
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College Algebra & Trigonometry
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c○2018 Richard W. Beveridge This
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College Algebra & Trigonometry, 2018a

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