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

96010Sin 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?

bh 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 25Sino

1-

Error → No such triangle