The document discusses using the Law of Sines to determine the number of possible solutions to a triangle based on the relationships between sides and sines of angles. It states that if a is less than b sin A, there is no solution; if a equals b sin A, there is one solution; and if a is greater than b sin A, there are two solutions. It also provides instructions for solving triangles using the Law of Sines when there is one or two possible solutions.

1. Ambiguous Case for
Law of Sines
Precal
5-7
p. 320

2. a < b
a < b sinA
No sol.
a = b sinA
One sol.
a > b sin A
Two sol.
a ≥ b
One solution

3. a < b
a < b sinA
No sol.
a = b sinA
One sol.
a > b sin A
Two sol.
a ≥ b
One solution

4. a < b
a < b sinA
No sol.
a = b sinA
One sol.
a > b sin A
Two sol.
a ≥ b
One solution

5. a < b
a < b sinA
No sol.
a = b sinA
One sol.
a > b sin A
Two sol.
a ≥ b
One solution

6. a ≤ b
No solution
a > b
One solution

7. a ≤ b
No solution
a > b
One solution

8. Solving Triangles
• Find the number of possible solutions--
none, one, or two.

9. Solving Triangles
• Find the number of possible solutions--
none, one, or two.
• If one--use Law of Sines to solve the
triangle.

10. Solving Triangles
• Find the number of possible solutions--
none, one, or two.
• If two--use Law of Sines to find one angle,
then find another angle less than 180°
with the same sine value.