Law of Sines

Section 6.1

• So far we have learned how to solve for only one type of triangle

• Right Triangles

• Next, we are going to be solving oblique triangles

• Any triangle that is not a right triangle

In general:

C

cA

a

B

b

• To solve an oblique triangle, we must know 3 pieces of information:

a) 1 Side of the triangle

b) Any 2 other componentsa) Either 2 sides, an angle and a side, and 2 angles

• AAS• ASA• SSA• SSS• SAS

C

cA

a

B

bLaw of Sines

Law of Sines

• If ABC is a triangle with sides a, b, and c, then:

CSin

c

BSin

b

ASin

a

ASA or AAS

A

C

B27.4 102.3º

28.7º

A = a = c =

49º

28.7Sin

27.4

49Sin

a

49Sin 27.4 28.7aSin

28.7Sin

4927.4Sin a

43.06 a

43.06

ASA or AAS

A

C

B27.4 102.3º

28.7º

A = a = c =

49º

28.7Sin

27.4

102.3Sin

c

102.3Sin 27.4 28.7cSin

28.7Sin

102.327.4Sin c

55.75 c

43.0655.75

Solve the following Triangle:

• A = 123º, B = 41º, and a = 10

123º 41º

10

C

c

b

C = 16º

123º 41º

10

C

c

b

C = 16º

123Sin

10

41Sin

b

41Sin 10 123bSin

123Sin

4110Sin b

7.8 b

b = 7.8

123º 41º

10

C

c

b

C = 16º

123Sin

10

16Sin

c

16Sin 10 123cSin

123Sin

1610Sin c

3.3 c

b = 7.8c = 3.3

Solve the following Triangle:

• A = 60º, a = 9, and c = 10

60º

9C

10

b

How is this problem different?

B

What can we solve for?

60Sin

9

CSin

10

CSin 9 6010Sin

9

6010Sin CSin

o74.2 C

60º

9

C

10

b

B

C = 74.2º

60Sin

9

45.8Sin

b

45.8Sin 9 60bSin

60Sin

45.89Sin b

7.5 C

60º

9

C

10

b

B

C = 74.2ºB = 45.8ºc = 7.5

What we covered:

• Solving right triangles using the Law of Sines when given:

1) Two angles and a side (ASA or AAS)2) One side and two angles (SSA)

• Tomorrow we will continue with SSA

SSA

The Ambiguous Case

Yesterday• Yesterday we used the Law of Sines to solve

problems that had two angles as part of the given information.

• When we are given SSA, there are 3 possible situations.

1) No such triangle exists2) One triangle exists3) Two triangles exist

Consider if you are given a, b, and A

A

ab h

Can we solve for h?

b

h A Sin

h = b Sin A

If a < h, no such triangle exists

Consider if you are given a, b, and A

A

ab h

If a = h, one triangle exists

Consider if you are given a, b, and A

A

ab h

If a > h, one triangle exists

Consider if you are given a, b, and A

A

ab

If a ≤ b, no such triangle exists

Consider if you are given a, b, and A

A

ab

If a > b, one such triangle exists

Hint, hint, hint…

• Assume that there are two triangles unless you are proven otherwise.

Two Solutions

• Solve the following triangle.

a = 12, b = 31, A = 20.5º

20.5º

31 12

2 Solutions

First Triangle

• B = 64.8º• C = 94.7º• c = 34.15

Second Triangle

• B’ = 180 – B = 115.2º• C’ = 44.3º• C’ = 23.93

Problems with SSA

1) Solve the first triangle (if possible)2) Subtract the first angle you found from 1803) Find the next angle knowing the sum of all

three angles equals 1804) Find the missing side using the angle you

found in step 3.

A = 60º; a = 9, c = 10

First Triangle

• C = 74.2º• B = 48.8º• b = 7.5

Second Triangle

• C’ = 105.8º• B’ = 14.2º• b ’ = 2.6

One Solution

• Solve the following triangle. What happens when you try to solve for the second triangle?

a = 22; b = 12; A = 42º

a = 22; b = 12; A = 42º

First Triangle

• B = 21.4º• C = 116.6º• c = 29.4

Second Triangle

• B’ = 158.6º• C’ = -20.6º

No Solution

• Solve the following triangle.a = 15; b = 25; A = 85º

15

85Sin 25Sin

o1-

Error → No such triangle

Law of Sines

Section 6.1

Warm Up

• Solve the following triangles (if possible)

a) A = 58º; a = 20; c = 10

b) B = 78º; b = 207; c = 210

c) A = 62º; a = 10; b = 12

A = 58º; a = 20; c = 10

CSin

10

58Sin

20

CSin 20 5810Sin

)20

5810Sin (Sin C 1-

o25 C

C = 25º B = 97º

97Sin

b

58Sin

20

58Sin b 9720Sin

)58Sin

9720Sin ( b

23.4 b

b = 23.4

B = 78º; b = 207; c = 210

CSin

210

78Sin

207

CSin 207 78210Sin

)207

78210Sin (Sin C 1-

o82.9 C

C = 82.9º A = 19.1º

19.1Sin

a

78Sin

207

78Sin a 19.1207Sin

)78Sin

19.1207Sin ( a

69.2 a

b = 69.2

B = 78º; b = 207, c = 210

First Triangle

• C = 82.9º• A = 19.1º• a = 69.2

Second Triangle

• C’ = 97.1º• A’ = 4.9º• a ’ = 18.1

A = 62º; a = 10; b = 12

BSin

12

62Sin

10

62Sin 12 B10Sin

)10

6212Sin (Sin C 1-

No such triangle

Area of an oblique triangle

What is the area of a triangle?

A

ab hh = b Sin A

hb2

1 area

b = c

ASin cb2

1 area

Area of an Oblique Triangle

• The area of any triangle is one-half the product of the lengths of two sides times the sine of their included angle.

• i.e.: Area = ½ bc Sin A = ½ ac Sin b = ½ ab Sin B

Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102º.

Area = ½ (Side) (Side) Sine (angle)

Area = ½ (90) (52) Sine (102º)

Area = 1,189 m²

Find the area of the triangle:

• A = 35º, a = 14.7, b = 14.7

• The course for a boat race starts at a point A and proceeds in the direction S 52º W to point B, then in the direction of S 40º E to point C, and finally back to A. Point C lies 8 miles directly south of point A. How far was the race?

52

40

A

B

C

• An airplane left an airport and flew east for 169 miles. Then it turned northward to N 32º E. When it was 264 miles from the airport, there was an engine problem and it turned to take the shortest route back to the airport. Find θ, the angle through which the airplane turned.

A farmer has a triangular field with sides that are 450 feet, 900 feet, and an included angle of 30º. He wants to apply fall fertilizer to the field. If it takes one 40 pound bag of fertilizer to cover 6,000 square feet, how many bags does he need to cover the entire field?