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?
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