9.4 The Law of Cosines

Objective To use the law of cosines to find unknown parts of a triangle.

Data Required for Solving Oblique Triangles

Case 1 One side and two angles known: »SAA or ASA

Case 2 Two sides and one angle not included between the sides known: »SSA

This case may lead to more than one solution.

Ambiguous Case

Case 3 Two sides and one angle includedbetween the sides known:»SAS

Case 4 Three sides are known:»SSS

Solving an SAS Triangle

• The Law of Sines was good for– ASA - two angles and the included side – SAA - two angles and any side– SSA - two sides and an opposite angle

(being aware of possible ambiguity)

• Why would the Law of Sines not work for an SAS triangle?

15 12.526°No side opposite

from any angle to get the ratio

No side opposite from any angle to

get the ratio

Deriving the Law of Cosines

Write an equation using

Pythagorean theorem for

shaded triangle.

sin

cos

h b A

k b A

2 2 2

2 22

2 2 2 2 2 2

2 2 2 2 2

2 2 2

( )

sin cos

sin 2 cos cos

sin cos 2 cos

2 cos

a h c k

a b A c b A

a b A c c b A b A

a b A A c c b A

a b c c b A

b h a

k c - kA B

C

c

Deriving the Law of Cosines

Write an equation using

Pythagorean theorem for

shaded triangle.

sin( ) sin

cos( ) cos

h b A b A

k b A b A

2 2 2

2 22

2 2 2 2 2 2

2 2 2 2 2

2 2 2

( )

sin cos

sin 2 cos cos

sin cos 2 cos

2 cos

a h c k

a b A c b A

a b A c c b A b A

a b A A c c b A

a b c c b A

h b a

k cA B

C

c + k

- A

c + (– bcosA)

c – bcosA

Law of Cosines• Similarly

• Note the pattern

2 2 2

2 2 2

2 2 2

2 cos

2 cos

2 cos

a b c b c A

b a c a c B

c a b a b C

Include

Opposite

S A S

Use these to findmissing sides

2 2 2 2 cosa b c bc A 2 2 2 2 cosb a c ac B 2 2 2 2 cosc a b ab C

Law of Cosines

When C = 90o, cosC = 0, then the Law of Cosines reduces to c2 = a2 + b2, which indicates a right triangle.

When C < 90o, cosC > 0, then c2 < a2 + b2 by the amount 2abcosC.

When C > 90o, cosC < 0, then c2 > a2 + b2 by the amount –2abcosC.

If we solve the law of cosines for cos A, cos B, cos C, we obtain2 2 2

2 2 2

2 2 2

cos2

cos2

cos2

b c aA

bc

a c bB

ac

a b cC

ab

Law of Cosines

S S S

Use these to find missing angles

Choose the MethodSolving an Oblique Triangle

Given Begin by Using

Two angles and any side

(SAA or ASA)

Law of Sines

Two sides and an angle opposite one of them

(SSA) (Ambiguous Case)

Law of Sines

Two sides and their included angle

(SAS)

Law of Cosines

Three sides

(SSS)

Law of Cosines

Using the law of cosines, we can easily identify acute and obtuse angles.

The law of sines does not distinguish between obtuse and acute angles, however, because both types of angle have positive sine values.

The Law of Sines and Cosines

Applying the Cosine Law

• Now use it to solve the triangle we started with

• Label sides and angles– Side c first

15 12.526°

A B

C

c

2 2 2

2 2

2 cos

15 12.5 2 15 12.5 cos 26

c b a a b C

c

Case 3 Two sides and one angle includedbetween the sides known:»SAS

• Now calculate the angles– use b2 = a2 + c2 – 2ac cos B and solve for B.

15 12.526°

A B

C

c = 6.65

2 2 2 2 2 21cos cos

2 2

a c b a c bB B

a c a c

Applying the Cosine Law

93.75

93.75o

• The remaining angledetermined by subtraction A = 180o – 93.75o – 26o = 60.25o

60.25o

Example 1: The leading edge of each wing of the B-2 Stealth Bomber measures 105.6 feet in length. The angle between the wing's leading edges is 109.05o. What is the wing span (the distance from A to C)?

Applying the Cosine Law: Wing Span

A

C

105.6

o2 (105.6) sin(109.05 / 2)AC

[Solution 1] Not use the Law of Cosine

2 2 2 o

2 2 2 o

2 2 o

o

105.6 105.6 2 (105.6) (105.6) cos109.05

2 105.6 2 105.6 cos109.05

105.6 (2 2 cos109.05 )

105.6 2 2cos109.05

AC

AC

AC

AC

172 ft

[Solution 2] Use the Law of Cosine

172 ft

Try It OutDetermine the area of these triangles

127°12

24

76.3°

42.8°17.9a) b)

o1(12)(24)sin127

2K

o144sin127

115 u2

a

A 60.9o

o o

17.9

sin 60.9 sin 76.3

a

oo

17.9sin 60.9

sin 76.3a

o oo

1 17.9(17.9) sin 60.9 sin 42.8

2 sin 76.3K

97.9 u2

Cost of a LotExample 2: An industrial piece of real estate is priced at

$4.15 per square foot. Find the cost of a triangular lot measuring 324 feet by 516 feet by 412 feet.

516

412

324[Solution] Use the Law of Cosines to find the angle opposite to the shortest side.

A

2 2 2516 412 324 20689cos

2 516 412 26574A

22 20689

sin 1 cos 126574

A A

2781427550.62759

26574

1 278142755(516)(412) 66710.45

2 26574K

1 2781427554.15 (516)(412) 276848.37

2 26574Cost

Case 4 Three sides are known:» SSS

Example 3: The lengths of the sides of a triangle are 5, 10, and 12. Solve the triangle.

[Solution] Make a sketch

5

10 12

2 2 25 10 12cos .19

2 5 10

101.0

2 2 212 10 5cos 0.9125

2 12 10

24.1

180 101.0 24.1 54.9

Case 4 Three sides are known:•SSS

Applying the Cosine Law

Example 4: In the diagram at the right, AB = 5, BD = 2,

DC = 4, and CA = 7. Find AD.

[Solution] First we apply the law of cosines to ABC to find out the cosB:

2 2 25 6 7 12 1cos

2 5 6 60 5B

Then we apply the law of sines to ABD to find out AD:

2 2 2 15 2 2 5 2 25

5AD

Thus AD = 5.

19

20o

53

y2 = x2 + z2 – 2·x·z·cosY

192 = x2 + 532 – 2·x·53·cos20o

Law of Cosines - SSA

x

x2 – 106·cos 20o x + (532 – 192) = 0

2 2 2106cos 20 106cos 20 4(53 19 )

2x

o o

YZ

XExample 5: In XYZ, Y = 20o,

z = 53, y = 19. Find YZ or x.

Z

19

[Solution] This situation is SSA, which may have ambiguous cases. It seems we have to apply the law of sines. However, the question is only asks to find a side, YZ, or x. We then use the law of cosines. 44.1 55.5x x or

44.1 55.5

19

20o

53

x YZ

X

Z

19

44.1 55.5x x or

44.155.5

Law of Cosines - SSA

Example 5: In XYZ, Y = 20o,

z = 53, y = 19. Find YZ or x.

2 2 2106cos 20 106cos 20 4(53 19 )

2x

o o

Example 6: Find the three angles of the triangle.

C

BA

86

12

2 2 2cos

2a b cC

ab

Find the angle opposite the longest side first.

2 2 2122(

8 68 6)( )

64 36 14496

4496

117.3C

117.12

sin 3 in6

s BLaw of Sines: 26.4B

180 117.3 26.4A

36.3

117.3

26.4

36.3

[Solution]

Example 7: Solve the triangle.

67.8

759.9 6.2

sin sin A Law of Sines: 37.2A

180 75 37.2 67.8C

37.2

C

BA

6.2

759.52 2 2 2 cosb a c ac B

2 2( ) ( ) 2( )6.2 9.5 6.2 9.5( 7s 5)co

38.44 90.25 (117.8)(0.25882)

98.209.9b

9.9

Law of Cosines:

[Solution]

Example 8: ApplicationTwo ships leave a port at 9 A.M. One travels at a bearing of N 53 W at 12 mph, and the other travels at a bearing of S 67 W at 16 mph. How far apart will the ships be at noon?

53

67

c

36 mi

48 mi

C

At noon, the ships have traveled for 3 hours.

Angle C = 180 – 53 – 67 = 60

2 2 2 2 cosc a b ab C 2 2 2( )( ) c36 4 os 68 36 8 1 24 80 7

43 mic

The ships will be approximately 43 miles apart.

43 mi 60

N

Assignment

P. 352 #1 – 18, 22