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