103 Trigonometry Problems

1. Trigonometric Fundamentals 23
Existence, Uniqueness, and Trigonometric Substitutions
The fact that sin α = sin β, for α + β = 180 ◦ , has already helped us in many
places. It also helped us to explain why either side-side-angle (SSA) or area-sideside
information is not enough to determine the unique structure of a triangle.
Example 1.7. Let ABC be a triangle.
(a) Suppose that [ABC] =10 √ 3, |AB| =8, and |AC| =5. Find all possible
values of ̸ A.
(b) Suppose that |AB| =5 √ 2, |BC| =5 √ 3, and ̸
values of ̸ A.
C = 45 ◦ . Find all possible
(c) Suppose that |AB| =5 √ 2, |BC|=5, and ̸
of ̸ A.
C = 45 ◦ . Find all possible values
(d) Suppose that |AB| =5 √ 2, |BC| =10, and ̸
values of ̸ A.
(e) Suppose that |AB| =5 √ 2, |BC| =15, and ̸
values of ̸ A.
C = 45 ◦ . Find all possible
C = 45 ◦ . Find all possible
Solution:
(a) Note that b =|AC| =5, c =|AB| =8, and [ABC] =
2 1 bc sin A. Thus
√
sin A = 3
2 , and A = 60◦ or 120 ◦ (A 1 and A 2 in Figure 1.23).
(b) By the law of sines,wehave |BC|
120 ◦ .
(c) By the law of sines, we have |BC|
(A 3 in Figure 1.23)! (Why?)
sin A = |AB|
sin A = |AB|
√
sin C ,orsinA = 3
2 . Hence A = 60◦ or
sin C ,orsinA = 1 2 . Hence A = 30◦ only
(d) By the law of sines, we have sin A = 1, and so A = 90 ◦ .(A 4 in Figure 1.23)
(e) By the law of sines, we have sin A = 3 2
, which is impossible. We conclude
that there is no triangle satisfying the conditions of the problem.