19 may 2011
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19 May 2011. Algebra 2. Law of Cosines5/19. a 2 = b 2 + c 2 – 2bc* cos (A) b 2 = a 2 + c 2 – 2ac* cos (B) c 2 = a 2 + b 2 – 2ab* cos (C). Three forms of L.o.C . Law of Cosines5/19. Law of Sines is easier, but Law of Cosines can be used more often. - PowerPoint PPT PresentationTRANSCRIPT
19 May 2011
Algebra 2
Law of Cosines 5/19Three forms
of L.o.C.a2 = b2 + c2 – 2bc*cos(A)
b2 = a2 + c2 – 2ac*cos(B)
c2 = a2 + b2 – 2ab*cos(C)
Law of Cosines 5/19When to use
L.o.C.Law of Sines is easier, but Law of Cosines can be used more often.
Law of Cosines 5/19When to use
L.o.C.Law of Sines is easier, but Law of Cosines can be used more often
b = ?
Law of Cosines 5/19When to use
L.o.C.Law of Sines is easier, but Law of Cosines can be used more often
b = ?
5)()36(
7)( CSin
bSinASin
Law of Cosines 5/19When to use
L.o.C.Law of Sines is easier, but Law of Cosines can be used more often
b = ?
Law of Sines doesn’t help us!
5)()36(
7)( CSin
bSinASin
Law of Cosines 5/19When to use
L.o.C.Law of Sines is easier, but Law of Cosines can be used more often
b = ?
Law of Cosines to the rescue!!!
Law of Cosines 5/19When to use
L.o.C.Law of Sines is easier, but Law of Cosines can be used more often
b = ?
b2 = a2 + c2 – 2ac*cos(B)
Law of Cosines 5/19When to use
L.o.C.Law of Sines is easier, but Law of Cosines can be used more often
b = ?
b2 = a2 + c2 – 2ac*cos(B)
b2 = 72 + 52 – 2*7*5*cos(36)
Law of Cosines 5/19When to use
L.o.C.Law of Sines is easier, but Law of Cosines can be used more often
b = ?
b2 = a2 + c2 – 2ac*cos(B)
b2 = 72 + 52 – 2*5*7*cos(36)
b2 = 17.4
Law of Cosines 5/19When to use
L.o.C.Law of Sines is easier, but Law of Cosines can be used more often
b = ?
b2 = a2 + c2 – 2ac*cos(B)
b2 = 72 + 52 – 2*5*7*cos(36)
b2 = 17.4 b = 4.2
Law of Cosines 5/19When to use
L.o.C.Example 2: find c
C
15 26o
12.5 A c B
Which form of the L.o.C. should we use?
Law of Cosines 5/19When to use
L.o.C.Example 2: find c
C
15 26o
12.5 A c B
c2 = a2 + b2 – 2ab*cos(C)
Law of Cosines 5/19When to use
L.o.C.Example 2: find c
C
15 26o
12.5 A c B
c2 = a2 + b2 – 2ab*cos(C)
c2 = 12.52 + 152 – 2*12.5*15*cos(26)
Law of Cosines 5/19When to use
L.o.C.Example 2: find c
C
15 26o
12.5 A c B
c2 = a2 + b2 – 2ab*cos(C)
c2 = 12.52 + 152 – 2*12.5*15*cos(26)
c2 = 44.20
Law of Cosines 5/19When to use
L.o.C.Example 2: find c
C
15 26o
12.5 A c B
c2 = a2 + b2 – 2ab*cos(C)
c2 = 12.52 + 152 – 2*12.5*15*cos(26)
c2 = 44.2 c = 6.6
Student Practice:
• Complete #1,6,9,10 on pages 843,844• #1 find b• #6 find b• #9 find c• #10 find b
Law of Cosines 5/19Finding an
angleExample 3: find <B
C
18 9
A 10 B
Use L.o.C. formula that has <B
Law of Cosines 5/19Finding an
angleExample 3: find <B
C
18 9
A 10 B
b2 = a2 + c2 – 2ac*cos(B)
Law of Cosines 5/19Finding an
angleExample 3: find <B
C
18 9
A 10 B
b2 = a2 + c2 – 2ac*cos(B)
182 = 92 + 102 – 2*9*10*cos(B)
Law of Cosines 5/19Finding an
angleExample 3: find <B
C
18 9
A 10 B
b2 = a2 + c2 – 2ac*cos(B)
182 = 92 + 102 – 2*9*10*cos(B)
182 – 92 – 102= -2*9*10*cos(B)
Law of Cosines 5/19Finding an
angleExample 3: find <B
C
18 9
A 10 B
b2 = a2 + c2 – 2ac*cos(B)
182 = 92 + 102 – 2*9*10*cos(B)
182 – 92 – 102= -2*9*10*cos(B) = cos(B)10*9*2-
10 - 9 - 18 222
Law of Cosines 5/19Finding an
angleExample 3: find <B
C
18 9
A 10 B
b2 = a2 + c2 – 2ac*cos(B)
182 = 92 + 102 – 2*9*10*cos(B)
182 – 92 – 102= -2*9*10*cos(B) = cos(B) cos(B) = -.79410*9*2-
10 - 9 - 18 222
Law of Cosines 5/19Finding an
angleExample 3: find <B
C
18 9
A 10 B
b2 = a2 + c2 – 2ac*cos(B)
182 = 92 + 102 – 2*9*10*cos(B)
182 – 92 – 102= -2*9*10*cos(B) = cos(B) <B = 142o10*9*2-
10 - 9 - 18 222
Student Practice
• Complete # 2,11,12,19 on pages 843-844• Always find the angle opposite the longest side
Law of Cosines 5/19Using L.o.C. You will be asked to “solve” a triangle.
This means find all missing sides and angles.
Easiest way to do this is to use the Law of Cosines once, then use the Law of Sines.
Law of Cosines 5/19Using L.o.C. Solve the triangle
Step 1: Use Law of Cosines to find the largest angle (A)
Law of Cosines 5/19Using L.o.C. Solve the triangle
Step 1: Use Law of Cosines to find the largest angle (A)
“Mr. Meeks which formula do I use, there are three?”
Law of Cosines 5/19Using L.o.C. Solve the triangle
Step 1: Use Law of Cosines to find the largest angle (A)
Use the one that will give you < A.
Law of Cosines 5/19Using L.o.C. Solve the triangle
a2 = b2 + c2 – 2bc*cos(A)
Law of Cosines 5/19Using L.o.C. Solve the triangle
a2 = b2 + c2 – 2bc*cos(A)
162 = 92 + 102 – 2*9*10*cos(A)
Law of Cosines 5/19Using L.o.C. Solve the triangle
a2 = b2 + c2 – 2bc*cos(A)
162 = 92 + 102 – 2*9*10*cos(A)
162 - 92 - 102 = – 2*9*10*cos(A)
Law of Cosines 5/19Using L.o.C. Solve the triangle
a2 = b2 + c2 – 2bc*cos(A)
162 = 92 + 102 – 2*9*10*cos(A)
162 - 92 - 102 = – 2*9*10*cos(A)
= cos(A)10*9*2- 10 - 9 - 16 222
Law of Cosines 5/19Using L.o.C. Solve the triangle
a2 = b2 + c2 – 2bc*cos(A)
162 = 92 + 102 – 2*9*10*cos(A)
162 - 92 - 102 = – 2*9*10*cos(A)
= cos(A) - 0.4167 = cos(A)10*9*2- 10 - 9 - 16 222
Law of Cosines 5/19Using L.o.C. Solve the triangle
a2 = b2 + c2 – 2bc*cos(A)
162 = 92 + 102 – 2*9*10*cos(A)
162 - 92 - 102 = – 2*9*10*cos(A)
= cos(A) 115o = A10*9*2- 10 - 9 - 16 222
Law of Cosines 5/19Using L.o.C. Solve the triangle
Step 2: Use Law of Sines to find < B
Law of Cosines 5/19Using L.o.C. Solve the triangle
Step 2: Use Law of Sines to find < B
9)(
16)115( BSinSin
Law of Cosines 5/19Using L.o.C. Solve the triangle
Step 2: Use Law of Sines to find < B
sin(B) = 0.5098
9)(
16)115( BSinSin
Law of Cosines 5/19Using L.o.C. Solve the triangle
Step 2: Use Law of Sines to find < B
sin(B) = 0.5098 B = 31
9)(
16)115( BSinSin
Law of Cosines 5/19Using L.o.C. Solve the triangle
Step 3: Subtract to find C
180 – 115 – 31 = 34
C = 34o
VERY IMPORTANT
I just gave you an EXAMPLE of how to solve this.
Do NOT try to use steps 1-3 for every problem.
IT WILL NOT WORK.
Always think to yourself, “What do I need in order to be able to use the Law of Sines?”