Section 6.1.3 Day 1
The Ambiguous Case of the
Law of Sines
Lesson Objective:
Students will:
• Solve triangles that could have more than
one solution.
• Extend their understanding of the inverse
sine and cosine functions to triangles.
Recall
•The Law of Sines provides a
way to solve triangles that are
not right triangles
The Law of Sines is:
C
a
b
A
c
a
sin A
B

b
sinB

c
sinC
Recall
To use the Law of Sines, at
least one angle and the
measure of the side opposite
that angle must be known.
Law of Sines
If given:
•two angles and
•the side opposite one of the given
angles,
Then:
•one triangle exists and
•use Law of Sines to find missing
parts.
Only One Angle Given
When one angle and two sides
(with one side across from the
angle) are given, the following
may be true:
•no triangle exists
•one triangle exists
•two triangles exist
Case 1
Given the measure of angle A is less
than 90:
• If a = bsinA then one triangle exists.
• If a > bsinA and a > b one triangle
exists.
• If a > bsinA and a < b then 2
triangles exist.
• If a < bSinA then no triangle exists.
Case 2
Given the measure of angle A > 90:
• If a < b then no triangle exists.
• If a > b then one triangle exists.
Example 1
Solve
50sin60°
ABC if
50
A=60°
b=50 and
60°
A
a=33
C
33
a<bsinAno triangle
B
Example 2 C
Solve
ABC if
A=60°
b=50 and
a=45
50
A
45
60°
2 Triangles
B
Example 2 (cont.)
Solve
ABC if
A=60°
b=50 and
a=45
C
bsina
50 45
A
60°
B
a>bsinA;a<b2 triangles
Example 3
Solve
C
bsina
ABC if
50
A=60°
b=50 and
60°
A
a=65
1 Triangle
65
B
Example 3 (cont.)
Solve
C
bsina
ABC if
50
A=60°
b=50 and
60°
A
a=65
65
a>bsina & a>b1 triangle
B
Lesson Close
What is the purpose of the Law
of Sines?
Assignment
Board Problems #1-7