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6.1 - Law of Sines

PC 6.1 Notes

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Notes for the Law of Sines, including the Ambiguous case. Mr. Fjelstrom's BM 13.1.

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Page 1: PC 6.1 Notes

6.1 - Law of Sines

Page 2: PC 6.1 Notes

2

An oblique triangle is a triangle that has no right angles.

To solve an oblique triangle, you need to know the measure of at least one side and the measures of any

other two parts of the triangle – two sides, two angles, or one angle and one side.

C

BA

ab

c

Page 3: PC 6.1 Notes

3

Four Cases for solving a triangle…

Page 4: PC 6.1 Notes

3

Four Cases for solving a triangle…1. Two angles and any side (AAS or ASA)

A

Cc

A

Bc

Page 5: PC 6.1 Notes

3

Four Cases for solving a triangle…1. Two angles and any side (AAS or ASA)

2. Two sides and an angle opposite one of them (SSA)

A

Cc

A

Bc

C

c

a

Page 6: PC 6.1 Notes

3

Four Cases for solving a triangle…1. Two angles and any side (AAS or ASA)

2. Two sides and an angle opposite one of them (SSA)

3. Three sides (SSS)

A

Cc

A

Bc

a

cb

C

c

a

Page 7: PC 6.1 Notes

3

Four Cases for solving a triangle…1. Two angles and any side (AAS or ASA)

2. Two sides and an angle opposite one of them (SSA)

3. Three sides (SSS)

4. Two sides and their included angle (SAS)

A

Cc

A

Bc

a

cb

C

c

a

c

aB

Page 8: PC 6.1 Notes

4

Case 1 and Case 2:

Page 9: PC 6.1 Notes

4

Case 1 and Case 2: Law of SinesIf ABC is an oblique triangle with sides a, b, and c, then

Acute Triangle

C

BA

bh

c

a

C

BA

bh

c

a

Obtuse Triangle

Page 10: PC 6.1 Notes

5

Find the remaining angle and sides of the triangle.

Example 1 (ASA):

Page 11: PC 6.1 Notes

5

Find the remaining angle and sides of the triangle.

Example 1 (ASA):

C

BA

b

c

60°

10°

a = 4.5 ft

Page 12: PC 6.1 Notes

5

Find the remaining angle and sides of the triangle.

Example 1 (ASA):

The third angle in the triangle is A = 180° – A – B

= 180° – 10° – 60° = 110°

C

BA

b

c

60°

10°

a = 4.5 ft

110°

Page 13: PC 6.1 Notes

5

Find the remaining angle and sides of the triangle.

Example 1 (ASA):

The third angle in the triangle is A = 180° – A – B

= 180° – 10° – 60° = 110°

C

BA

b

c

60°

10°

a = 4.5 ft

110°

Use the Law of Sines to find side b and c.

Page 14: PC 6.1 Notes

5

Find the remaining angle and sides of the triangle.

Example 1 (ASA):

The third angle in the triangle is A = 180° – A – B

= 180° – 10° – 60° = 110°

C

BA

b

c

60°

10°

a = 4.5 ft

110°

Use the Law of Sines to find side b and c.

4.15 ft

Page 15: PC 6.1 Notes

5

Find the remaining angle and sides of the triangle.

Example 1 (ASA):

The third angle in the triangle is A = 180° – A – B

= 180° – 10° – 60° = 110°

C

BA

b

c

60°

10°

a = 4.5 ft

110°

Use the Law of Sines to find side b and c.

4.15 ft

0.83 ft

Page 16: PC 6.1 Notes

6

Watch out for SSA!

Page 17: PC 6.1 Notes

6

Watch out for SSA!SSA is not a congruency property. SO, if we are given two sides and the NON-included angle in a triangle, there are three possible scenarios…

Page 18: PC 6.1 Notes

6

Watch out for SSA!SSA is not a congruency property. SO, if we are given two sides and the NON-included angle in a triangle, there are three possible scenarios…

➡ Option 1: No triangle is formed

Page 19: PC 6.1 Notes

6

Watch out for SSA!SSA is not a congruency property. SO, if we are given two sides and the NON-included angle in a triangle, there are three possible scenarios…

➡ Option 1: No triangle is formed➡ Option 2: One triangle is formed

Page 20: PC 6.1 Notes

6

Watch out for SSA!SSA is not a congruency property. SO, if we are given two sides and the NON-included angle in a triangle, there are three possible scenarios…

➡ Option 1: No triangle is formed➡ Option 2: One triangle is formed➡ Option 3: Two triangles are formed

Page 21: PC 6.1 Notes

6

Watch out for SSA!SSA is not a congruency property. SO, if we are given two sides and the NON-included angle in a triangle, there are three possible scenarios…

➡ Option 1: No triangle is formed➡ Option 2: One triangle is formed➡ Option 3: Two triangles are formed

How is it determined? If …

Page 22: PC 6.1 Notes

6

Watch out for SSA!SSA is not a congruency property. SO, if we are given two sides and the NON-included angle in a triangle, there are three possible scenarios…

➡ Option 1: No triangle is formed➡ Option 2: One triangle is formed➡ Option 3: Two triangles are formed

How is it determined? If …opp ≥ adj

0 or 1 triangle

Page 23: PC 6.1 Notes

6

Watch out for SSA!SSA is not a congruency property. SO, if we are given two sides and the NON-included angle in a triangle, there are three possible scenarios…

➡ Option 1: No triangle is formed➡ Option 2: One triangle is formed➡ Option 3: Two triangles are formed

How is it determined? If …opp ≥ adj opp ≤ adj

0 or 1 triangle 0 or 2 triangles

Page 24: PC 6.1 Notes

7

Use the Law of Sines to solve the triangle.A = 110°, a = 125 inches, b = 100 inches

Example 2 (SSA):

Page 25: PC 6.1 Notes

7

Use the Law of Sines to solve the triangle.A = 110°, a = 125 inches, b = 100 inches

Example 2 (SSA):

C

BA

b = 100 in

c

a = 125 in

110°

Page 26: PC 6.1 Notes

7

Use the Law of Sines to solve the triangle.A = 110°, a = 125 inches, b = 100 inches

Example 2 (SSA):

C

BA

b = 100 in

c

a = 125 in

110°

There is either 0 or 1 triangle satisfying the given conditions

because opp ≥ adj

Page 27: PC 6.1 Notes

7

Use the Law of Sines to solve the triangle.A = 110°, a = 125 inches, b = 100 inches

Example 2 (SSA):

C

BA

b = 100 in

c

a = 125 in

110° 48.74°

There is either 0 or 1 triangle satisfying the given conditions

because opp ≥ adj

Page 28: PC 6.1 Notes

7

Use the Law of Sines to solve the triangle.A = 110°, a = 125 inches, b = 100 inches

Example 2 (SSA):

C ≈ 180° – 110° – 48.74°

C

BA

b = 100 in

c

a = 125 in

110° 48.74°

21.26°

= 21.26°

There is either 0 or 1 triangle satisfying the given conditions

because opp ≥ adj

Page 29: PC 6.1 Notes

7

Use the Law of Sines to solve the triangle.A = 110°, a = 125 inches, b = 100 inches

Example 2 (SSA):

C ≈ 180° – 110° – 48.74°

C

BA

b = 100 in

c

a = 125 in

110° 48.74°

21.26°

48.23 in

= 21.26°

There is either 0 or 1 triangle satisfying the given conditions

because opp ≥ adj

Page 30: PC 6.1 Notes

8

Use the Law of Sines to solve the triangle.A = 76°, a = 18 inches, b = 20 inches

Example 3 (SSA):

Page 31: PC 6.1 Notes

8

Use the Law of Sines to solve the triangle.A = 76°, a = 18 inches, b = 20 inches

Example 3 (SSA):

There is either 0 or 2 triangles satisfying the given conditions

because opp ≤ adj

Page 32: PC 6.1 Notes

8

Use the Law of Sines to solve the triangle.A = 76°, a = 18 inches, b = 20 inches

Example 3 (SSA):

There is no angle whose sine is 1.078.

There is either 0 or 2 triangles satisfying the given conditions

because opp ≤ adjC

AB

b = 20 ina = 18 in

76°

Page 33: PC 6.1 Notes

9

Use the Law of Sines to solve the triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA):

Page 34: PC 6.1 Notes

9

Use the Law of Sines to solve the triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA):

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

Page 35: PC 6.1 Notes

9

Use the Law of Sines to solve the triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA):

There is either 0 or 2 triangles that can be formed (opp ≤ adj)

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

Page 36: PC 6.1 Notes

9

Use the Law of Sines to solve the triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA):

72.2°

There is either 0 or 2 triangles that can be formed (opp ≤ adj)

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

Page 37: PC 6.1 Notes

9

Use the Law of Sines to solve the triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA):

72.2°

There is either 0 or 2 triangles that can be formed (opp ≤ adj)

49.8°

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

C ≈ 180° – 58° – 72.2° = 49.8°

Page 38: PC 6.1 Notes

9

Use the Law of Sines to solve the triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA):

72.2°

10.3 cm

There is either 0 or 2 triangles that can be formed (opp ≤ adj)

49.8°

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

Example continues …

C ≈ 180° – 58° – 72.2° = 49.8°

Page 39: PC 6.1 Notes

10

Use the Law of Sines to solve the second triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA) continued:

72.2°

10.3 cm

49.8°

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

a=11.4 cm

Page 40: PC 6.1 Notes

10

Use the Law of Sines to solve the second triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA) continued:

C

AB2

b = 12.8 cm

c

a = 11.4 cm58°

72.2°

10.3 cm

49.8°

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

a=11.4 cm

Page 41: PC 6.1 Notes

10

Use the Law of Sines to solve the second triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA) continued:

B2 ≈ 180° – 72.2° = 107.8 °

107.8°

C

AB2

b = 12.8 cm

c

a = 11.4 cm58°

72.2°

10.3 cm

49.8°

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

a=11.4 cm

Page 42: PC 6.1 Notes

10

Use the Law of Sines to solve the second triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA) continued:

B2 ≈ 180° – 72.2° = 107.8 °

107.8°

C

AB2

b = 12.8 cm

c

a = 11.4 cm58°

14.2°

72.2°

10.3 cm

49.8°

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

C ≈ 180° – 58° – 107.8° = 14.2°

a=11.4 cm

Page 43: PC 6.1 Notes

10

Use the Law of Sines to solve the second triangle.A = 58°, a = 11.4 cm, b = 12.8 cm

Example 4 (SSA) continued:

B2 ≈ 180° – 72.2° = 107.8 °

107.8°

C

AB2

b = 12.8 cm

c

a = 11.4 cm58°

14.2°

3.3 cm

72.2°

10.3 cm

49.8°

a = 11.4 cm

C

AB1

b = 12.8 cm

c58°

C ≈ 180° – 58° – 107.8° = 14.2°

a=11.4 cm

Page 44: PC 6.1 Notes

11

Area of an Oblique Triangle

Page 45: PC 6.1 Notes

11

Area of an Oblique Triangle

Page 46: PC 6.1 Notes

11

Area of an Oblique Triangle

Find the area of the triangle.A = 74°, b = 103 inches, c = 58 inches

Example 5:

Page 47: PC 6.1 Notes

11

Area of an Oblique Triangle

C

BA

b

c

aFind the area of the triangle.A = 74°, b = 103 inches, c = 58 inches

Example 5:

74°

103 in

58 in

Page 48: PC 6.1 Notes

11

Area of an Oblique Triangle

C

BA

b

c

aFind the area of the triangle.A = 74°, b = 103 inches, c = 58 inches

Example 5:

74°

103 in

58 in

Page 49: PC 6.1 Notes

11

Area of an Oblique Triangle

C

BA

b

c

aFind the area of the triangle.A = 74°, b = 103 inches, c = 58 inches

Example 5:

74°

103 in

58 in

Page 50: PC 6.1 Notes

11

Area of an Oblique Triangle

C

BA

b

c

aFind the area of the triangle.A = 74°, b = 103 inches, c = 58 inches

Example 5:

74°

103 in

58 in

Page 51: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

Page 52: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

Page 53: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

14°

Page 54: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

Flagpole height: b

14°

Page 55: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

Flagpole height: b

14°

Page 56: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

Flagpole height: b

14°

Page 57: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

Flagpole height: b

16 m14°

Page 58: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

Flagpole height: b

16 m14°

Page 59: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

20°

Flagpole height: b

16 m14°

Page 60: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

20°

Flagpole height: b70°

16 m14°

Page 61: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

20°

Flagpole height: b70°

34°

16 m14°

Page 62: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

20°

Flagpole height: b70°

34°

16 m14°

B

A

C

Page 63: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

20°

Flagpole height: b70°

34°

16 m14°

B

A

C

Page 64: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

20°

Flagpole height: b70°

34°

16 m14°

B

A

C

Page 65: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

20°

Flagpole height: b70°

34°

16 m14°

B

A

C

Page 66: PC 6.1 Notes

12

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14° with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20°. How tall is the flagpole?

Application:

20°

Flagpole height: b70°

34°

16 m14°

The flagpole is approximately 9.5 meters tall.

B

A

C

Page 67: PC 6.1 Notes

Homework:

Assignment #2 p. 436 #7, 8, 13, 19-22, 29-31, 35, 36, 39, 40, 42