College Algebra & Trigonometry, 2018a

11.2. THE LAW OF SINES: THE AMBIGUOUS CASE 481
45
sin 112
=
c
◦ sin 38.4 ◦
45
0.9272 ≈ c
0.6211
0.6211∗ 45
0.9272 ≈ c
30.1 ≈ c
∠A = 112 ◦ a =45
∠B ≈ 29.6 ◦ b =24
∠C ≈ 38.4 ◦ c ≈ 30.1
Example 2
Solve the triangle if: ∠A =38 ◦ , a =40, b =52
Round the angles and side lengths to the nearest 10 th .
Using the Law of Sines, we can say that:
sin 38 ◦
40
= sin B
52
0.6157
40
≈ sin B
52
52∗ 0.6157
40
≈ sin B
0.8004 ≈ sin B
Just as in the previous example, we can find sin −1 (0.8004) ≈ 53.2 ◦ . But again,
there is a Quadrant II angle whose sine has the same value ≈ 0.8004. The angle
126.8 ◦ has a sine ≈ 0.8004 and a reference angle of 53.2 ◦ . With ∠A =38 ◦ , both of
these angles (53.2 ◦ and 126.8 ◦ ) could potentially fit in the triangle with angle A.