College Algebra & Trigonometry, 2018a

492 CHAPTER 11. THE LAW OF SINES THE LAW OF COSINES
The calculation for this problem is slightly different from the last one because the
cosine of 110 ◦ will be negative:
a 2 = b 2 + c 2 − 2bc cos A
a 2 =35 2 +30 2 − 2 ∗ 35 ∗ 30 ∗ cos 110 ◦
a 2 ≈ 1225 + 900 − 2100 ∗ (−0.3420)
a 2 ≈ 2125 + 718.2
a 2 ≈ 2843.2
a ≈ 53.3
In this problem, since we were given an obtuse angle, then the other two angles
must be acute and we don’t have to worry about the ambiguous case in using the
Law of Sines.
sin 110 ◦
53.3 =sin B
35
sin 110 ◦
53.3 =sin C
30
35∗ 0.9397
53.3
=sinB 30∗0.9397
53.3 =sinC
0.61706 ≈ sin B 0.5289 ≈ sin C
38.1 ◦ ≈ B 31.9 ◦ ≈ C
So the angles and sides of the triangle would be:
∠A = 110 ◦ a ≈ 53.3
∠B ≈ 38.1 ◦ b =35
∠C ≈ 31.9 ◦ c =30