1. Applications of Trigonometric Functions
Chapter 7
Applications of Trigonometric Functions
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2. The Law of Sines: Ambiguous Case
SECTION 7.2
Discuss the ambiguous case.
Solve possible SSA triangles. 1 2

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THE AMBIGUOUS CASE
If we know the lengths of two sides of a triangle and the measure of the angle opposite one of these sides, we can have a result that is
(1) not a triangle,
(2) one triangle, or
(3) two different triangles.
For this reason, Case 2 (SSA triangles) is called the ambiguous case.
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THE AMBIGUOUS CASE
A is an acute angle.
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THE AMBIGUOUS CASE
A is an acute angle. ≥
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THE AMBIGUOUS CASE
A is an obtuse angle.
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THE AMBIGUOUS CASE
If A is an acute angle:
Case
Conditions on a
Number of Triangles 1
a < b sin A
None 2
a = b sin A
One right triangle 3
b sin A < a < b
Two 4
a ≥ b
One
If A is an obtuse angle:
Case
Conditions on a
Number of Triangles 1
a ≤ b
None 2
a > b
One
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EXAMPLE 3
Solving the SSA Triangles
Solve triangle ABC with B = 32°, b = 100 feet, and c = 150 feet. Round each answer to the nearest tenth.
Solution
Step 1
Make a chart of the six parts, the known and unknown parts.
A = ?
a = ?
B = 32°
b = 100
C = ?
c = 150
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EXAMPLE 3
Solving the SSA Triangles
Solution continued
Step 2
Apply the Law of Sines to the two ratios in which three of the four quantities are known. Solve for the fourth quantity. Use the form of the Law of Sines in which the unknown quantity is in the numerator.
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EXAMPLE 3
Solving the SSA Triangles
Solution continued
Step 3
If the sine of the angle, say θ, in Step 2 is greater than 1, there is no triangle with the given measurements. If sin θ is between 0 and 1, go to Step 4.
sin C ≈ < 1, so go to Step 4.
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EXAMPLE 3
Solving the SSA Triangles
Solution continued
Step 4
Let sin θ be x, with 0 < x ≤ 1. If x ≠ 1, then θ has two possible values:
θ1 = sin–1x with 0 < θ1 <90°
θ2 = 180° – θ1
Two possible values of C follow:
C1 ≈ sin–1(0.7949) ≈ 52.6°
C2 ≈ 180° – 52.6° = 127.4°
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EXAMPLE 3
Solving the SSA Triangles
Solution continued
Step 5
If (given angle) + θ1 ≥ 180°, there is no triangle. If x ≠ 1, with (given angle) + θ2 < 180°, then there are two triangles. Otherwise, there is only one triangle.
B + C1 = 32° ° = 84.6° < 180°
B + C2 = 32° ° = 159.4° < 180°
We have two triangles with the given measurements.
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EXAMPLE 3
Solving the SSA Triangles
Solution continued
Step 6
Find the third angle of the triangle(s).
BAC1 ≈ 180° – 32° – 52.6° = 95.4°
BAC2 ≈ 180° – 32° – 127.4° = 20.6°
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EXAMPLE 3
Solving the SSA Triangles
Solution continued
Step 7
Use the Law of Sines to find the remaining side(s).
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EXAMPLE 3
Solving the SSA Triangles
Solution continued
Step 8
Show the solution(s).
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EXAMPLE 4
Solving an SSA Triangle (No Solution)
Solve triangle ABC with A = 50º, a = 8 inches, and b = 15 inches.
Solution
Step 1 Make a chart.
A = 50°
a = 8
B = ?
b = 15
C = ?
c = ?
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EXAMPLE 4
Solving an SSA Triangle (No Solution)
Solution continued
Step 2 Apply the Law of Sines.
Step 3
Because sin B ≈ 1.44 > 1, no triangle has the given measures.
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EXAMPLE 5
Solving an SSA Triangle (One Solution)
Solve triangle ABC with C = 40º, c = 20 meters, and a = 15 meters. Round each answer to the nearest tenth.
Solution
Step 1 Make a chart.
A = ?
a = 15
B = ?
b = ?
C = 40°
c = 20
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EXAMPLE 5
Solving an SSA Triangle (One Solution)
Solution continued
Step 2 Apply the Law of Sines.
Step 3
Since sin A is between 0 and 1, go to Step 4.
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EXAMPLE 5
Solving an SSA Triangle (One Solution)
Solution continued
Step 4
A = sin–1(0.4821) ≈ 28.8°. Two possible values of A are A1 ≈ 28.8° and A2 ≈ 180° –28.8° = 151.2°.
Step 5
Because C + A2 = 40° ° = 191.2° > 180°, there is no triangle with vertex A2. Thus, only one triangle has measure of angle A1 ≈ 28.8°.
Step 6
The third angle at B has measure ≈ 180° – 40° – 28.8° = 111.2°.
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EXAMPLE 5
Solving an SSA Triangle (One Solution)
Solution continued
Step 7 Find the remaining side length.
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EXAMPLE 5
Solving an SSA Triangle (One Solution)
Solution continued
Step 8 Show the solution.
A1 ≈ 28.8°
a = 15 meters
B ≈ 111.2°
b = 29.0 meters
C = 40°
c = 20 meters
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