College Algebra & Trigonometry, 2018a

11.1. THE LAW OF SINES 475
This will make our value for side a more accurate:
c
sin C = a
sin A
5
sin 95
=
a
◦ sin 40 ◦
Multiplying on both sides by sin 40 ◦ , then
sin 40 ◦ 5
∗
sin 95
= a ◦
To arrive at an approximate value for sin 40 ◦ 5
∗
sin 95
, we can say:
◦
5
0.6428∗
0.9962 ≈ a
3.23 ≈ a
∠A =40 ◦ a ≈ 3.23
∠B =45 ◦ b ≈ 3.55
∠C =95 ◦ c =5
Example 2
Some problems don’t come with diagrams:
Solve the triangle if: ∠A =40 ◦ , ∠B =20 ◦ , a =2.
Round side lengths to the nearest 100 th .
Just as in the previous example, we can begin by finding the measure of the third
angle ∠C. This would be 180 ◦ − (40 ◦ +20 ◦ ) = 180 ◦ − 60 ◦ = 120 ◦ = ∠C
To find the missing sides, we should use the complete side-angle pair that is given
in the problem: ∠A =40 ◦ and a =2.