The Law of cosines

THE LAW OF COSINESOur last new Section…………5.6

Deriving the Law of Cosines

a

c

b

A B(c,0)

C(x, y)

a

c

b

A B(c,0)

C(x, y)

a

c

b

A B(c,0)

C(x, y)In all three cases:

cos xAb

sin yAb

Rewrite:cosx b A siny b A

Deriving the Law of Cosines

a

c

b

A B(c,0)

C(x, y)

a

c

b

A B(c,0)

C(x, y)

a

c

b

A B(c,0)

C(x, y) Set a equal to the distance from C to Busing the distance formula:

2 20a x c y

Deriving the Law of Cosines

2 20a x c y

cosx b A siny b A

22 2a x c y

2 2cos sinb A c b A 2 2 2 2 2cos 2 cos sinb A bc A c b A

2 2 2 2cos sin 2 cosb A A c bc A 2 2 2 cosb c bc A

Law of Cosines

2 2 2 2 cosa b c bc A

Let ABC be any triangle with sides and angleslabeled in the usual way. Then

2 2 2 2 cosb a c ac B 2 2 2 2 cosc a b ab C

Note: While the Law of Sines was used to solve AAS andASA cases, the Law of Cosines is required for SAS and SSScases. Either method can be used in the SSA case, butremember that there might be 0, 1, or 2 triangles.

Guided Practice11a

Solve ABC, given the following.5b 20C

A

B

C

205

11

c

6.529

2 2 2 2 cosc a b ab C 2 211 5 2 11 5 cos20c

2 2 2 2 cosa b c bc A

144.8 2 2 2

1 11 5cos2 5

cAc

180 15.2B A C

Guided Practice9a

Solve ABC, given the following.7b 5c

A

BC

7

9

95.7

2 2 29 7 5 2 7 5 cos A 5

180 33.6C A B

2 2 2

1 9 7 5cos2 7 5

A

50.7

2 2 27 9 5 2 9 5 cosB

2 2 2

1 7 9 5cos2 9 5

B

Recall some diagrams :

a

c

b

A B(c,0)

C(x, y)

a

c

b

A B(c,0)

C(x, y)

a

c

b

A B(c,0)

C(x, y)For all three triangles: sin A

yb

Rewrite: siny b AThis can be considered theheight of each triangle, whileside c would be the base…

Area of a TriangleArea = (base)(height)

1 1 1sin sin sin2 2 2bc A ac B ab C

Generalizing:

Note: These formulas work in SAS cases…

12

= (c)(b sinA)12

= bc sinA12

Area =

Theorem: Heron’s FormulaLet a, b, and c be the sides of ABC, and let s denote

s s a s b s c

Clearly, this theorem is used in the SSS case…

the semiperimeter (a + b + c)/2. Then the area of

ABC is given by:

Area =

Guided PracticeFind the area of the triangle described.

52A An SAS case!!!(a)

Area =

14mb 21mc 1 sin2bc A 1 14 21 sin 52

2

2115.838m

112C An SAS case!!!(b)

Area =

1.8ina 5.1inb1 sin2ab C 1 1.8 5.1 sin112

2

24.256in

Guided PracticeDecide whether a triangle can be formed with the given sidelengths. If so, use Heron’s formula to find the area of the triangle.

5a Yes the sum of any twosides is greater than thethird side!!!

(a) 9b 7c

2a b cs

Can a triangle be formed?

Find the semiperimeter:

5 9 72

21 10.52

10.5 10.5 5 10.5 9 10.5 7A Heron’s formula:

17.412

Guided PracticeDecide whether a triangle can be formed with the given sidelengths. If so, use Heron’s formula to find the area of the triangle.

7.2a No b + c < a

(b) 4.5b 2.5c Can a triangle be formed?

Guided PracticeDecide whether a triangle can be formed with the given sidelengths. If so, use Heron’s formula to find the area of the triangle.

18.2a Yes the sum of any twosides is greater than thethird side!!!

(c) 17.1b 12.3c

2a b cs

Can a triangle be formed?

Find the semiperimeter:

18.2 17.1 12.32

23.8

23.8 23.8 18.2 23.8 17.1 23.8 12.3A Heron’s formula:

101.337

Guided Practicep.494: #36

(a) Find the distance from the pitcher’s rubber to the far cornerof second base. How does this distance compare with the distance from the pitcher’s rubber to first base?

90 2

A90 2 60.5(First

base)

Second base

(Pitcher’srubber) B

(Home plate)C

90 ft60.5 ft

45

c

The home-to-second segment is thehypotenuse of a right triangle, whichhas a length of…

Distance from pitcher to second:

66.779ft

Guided Practicep.494: #36

(a) Find the distance from the pitcher’s rubber to the far cornerof second base. How does this distance compare with the distance from the pitcher’s rubber to first base?

2 260.5 90 2 60.5 90 cos 45c

A

(Firstbase)

Second base

(Pitcher’srubber) B

(Home plate)C

90 ft60.5 ft

45

c

Solve for c with the Law of Cosines:

63.717ft66.779 ft

Guided Practicep.494: #36

(b) Find angle B in triangle ABC.

2 2 290 60.5 2 60.5 cosc c B

A

(Firstbase)

Second base

(Pitcher’srubber) B

(Home plate)C

90 ft60.5 ft

45

c

Again, the Law of Cosines:

92.8

2 2 21 90 60.5cos

2 60.5cBc

66.779 ft

Whiteboard Practice51A

Solve ABC, given the following.

11a 12bC

BA

12

c13.385,1.718c

2 2 211 12 2 12 cos51c c

11

58.0

2 24cos51 23 0c c

2 2 2

1 11

1

12 11cos2 11

cBc

1 13.385c

2 2 21 1 112 11 2 11 cosc c B

51

1 1180C A B 71.0

Whiteboard Practice51A

Solve ABC, given the following.

11a 12bC

BA

12

c

11

122.0

2 2 21 2

22

12 11cos2 11

cBc

2 1.718c

2 2 22 2 212 11 2 11 cosc c B

51

2 2180C A B 7.0

13.385,1.718c

2 2 211 12 2 12 cos51c c

2 24cos51 23 0c c

Whiteboard Practice74A

Solve ABC, given the following.

6.1a 8.9bC

BA

8.9

c

2 2 26.1 8.9 2 8.9 cos74c c 6.1 2 17.8cos74 42 0c c

74 No real solutions!!! No triangle is formed!!!