6.2 Law of Cosines

What would happen if we were given:

a = 6, b = 4, C = 60º

6Sin A

= 4Sin B = c

Sin 60º

What would happen if we were given:

a = 4, b = 5, c = 3

4Sin A

= 5Sin B = 3

Sin C

Law of Cosines Use Law of Cosines when the given

information is:

a) SSS (three lowercase letters)

b) SAS (2 sides and an angle, all 3 letters)

Law of Cosines

a2= b

2 + c – 2bc Cos A2

b2= a

2 + c – 2ac Cos B2

c2= a

2 + b – 2ab Cos C2

Alternative Form Solve these three equations for:

Cos A

Cos B

= b + c - a2bc

2 2 2

a + c - b2ac

2 2 2=

a + b - c2ab

2 2 2Cos C=

a = 8; b = 19; c = 14 (SSS)

→ Find the largest angle first

a + c - b2ac

2 2 2Cos B= =

8 + 14 - 192(8)(14)

2 2 2

B = 116.8º

→ remember to use the inverse Cosine to find an angle

a = 8; b = 19; c = 14; B = 116.8º

Once you know one angle, it is easiest to now use the Law of Sines.

8Sin A

= 19Sin 116.8º = 14

Sin C

A = 22.08º C = 41.12º

Solve the following triangles:

a)a = 5; b = 8; c = 9

b)a = 9; b = 7; c = 10

a = 5; b = 8; c = 9

a + b - c2ab

2 2 2Cos C= =

5 + 8 - 92(5)(8)

2 2 2

C = 84.3º

5Sin A

9Sin 84.3º=

A = 33.6º B = 62.2º

a = 10; b = 7; c = 9

b + c - a2bc

2 2 2Cos A= =

7 + 9 - 102(7)(9)

2 2 2

A = 76.2º

10Sin 76.2º

7Sin B=

B = 42.8º C = 61º

A = 115º; b = 15; c = 10 (SAS)

What do we have enough information to solve for?

a2= b

2 + c – 2bc Cos A2

a2= 15

2+ 10 – 2(15) (10) Cos 115º

2

a = 451.785482 a = 21.3

A = 115º; b = 15; c = 10 Now use the Law of Sines

21.3Sin 115º

= 15Sin B = 10

Sin C

B = 39.7º C = 25.2º

a = 21.3

Solve the following triangles:

a)a = 6; b = 4; C = 60º

b)a = 3; c = 2; B = 110º

a = 6; b = 4; C = 60º

c2= a

2 + b – 2ab Cos C2

c = 5.29

4Sin B

= 5.29Sin 60º

B = 40.9ºA = 79.1º

a = 3; c = 2; B = 110º

b2= a

2 + c – 2ac Cos B2

b = 4.14

2Sin C

= 4.14Sin 110º

C = 27ºA = 43º

Find the missing information:

5

8

c

d45º

β

d2

= 5 + 8 – 2(5) (8) Cos 45º22

d = 5.69

Find the missing information:

5

8

c

d45º

β8

135º

c2

= 5 + 8 – 2(5) (8) Cos 135º22

d = 12.07

Find the missing information:

25

35

c

dθ

120º

c2

= 25+ 35 – 2(25) (35) Cos 120º22

c = 52.2

35

Find the missing information:

25

35

c

d60º

8120º

d2

= 25+ 35 – 2(25) (35) Cos 60º22

d = 31.2

θ

6.2 Law of CosinesHeron’s Area Formula

Heron’s Area Formula Any triangle with given sides of lengths

a, b, and c, has an area of:

Area = s (s – a) (s – b) (s – c)

where s = a + b + c2

Find the area of a triangle have sides of lengths a = 43 meters, b = 53 meters, and c = 72 meters.

a = 43 b = 53 c = 72

s =

Area = 84 (84 – 43) (84 – 53) (84 – 72)

43 + 53 + 722

= 84

= 1131.89 square meters

a = 5 b = 7 c = 10

s =

Area = 11 (11 – 5) (11 – 7) (11 – 10)

5 + 7 + 102

= 11

= 16.25 square inches

7 in.5 in.

10 in.

A radio tower 500 feet high is located on the side of a hill with an inclination to the horizontal of 5º. How long should two guy wires be if they are to connect to the top of the tower and be secured at two points 100 feet directly above and directly below the base of the tower?

5º

100 100