6.2 Law of Cosines

To solve an oblique triangle where you are given the measurement of the three sides (SSS) or two sides and the included angle (SAS), you would use the Law of Cosines:

Law of Cosines:

a

b

c

Example 1: given SSS - find the three angles of the triangle given a = 9, b = 3 and c = 11

a

9

81 = 9 + 121 - (66) cos A

81 - 9 - 121 = - 66 cos A

-49 = -66 cos A

49/66 = Cos A

A = 42.06166467°

b

3

9 = 81 + 121 - 198 cos B

-193 = -198 cos B

193/198 = cos B

B = 12.90352054°

c

11

121 = 81 + 9 - 54 cos C

31 = -54 cos C

-31/54 = cos C

C = 125.0348148°

Example 2: Two Sides and the Included Angle - SAS

C = 108° , a = 10, b = 6.5

c

c

c

c = 13.50637662

Now find the other 2 angles:

a

10

100 = 224.64722093 - 175.582896 cos A

-124.6722093 = - 175.582896 cos A

.7100475736 = cos A

A = 44.76121382°

B = 180° - 108° - 44.76121382° = 27.23878618°

Example 3: If a street light is 2 feet in length on the post, the top is 3 feet and the distance from the bulb to the post is 4.5 feet. What is the angle measurement from the post to the top of the sign?

4.5

20.25 = 4 + 9 - 12 Cos A

7.25 = -12 Cos A

-7.25/12 = Cos A

A = 127.1688997°

so therefore, the angle from the post to the top of the sign would be

180° - 127.1688997° = 52.83110034°

Heron’s Area Formula - given any triangle with sides of lengths a, b, and c, the area of the triangle is:

Example 4: The Landau Building in Cambridge, Massachusetts, is a building with a triangular-shaped base. The lengths of the sides of the triangular base are 145 feet, 257 feet, and 290 feet. Find the area of the base of the building.

s = (145 + 257 + 290)/2 = 692/2 = 346 feet

Area = √(346(346 - 145)(346 - 257)(346 - 290))

= 18617.66 square feet

OR

Using Area = ½ bc Sin A, first find the measure of angle A

a

145

21025 = 66049 + 84100 - 149060 Cos A

-129124 = - 149060 cos A

129124/149060 = cos A

A = 29.97365689°

Area = ½ (257)(290) sin 29.97365689°

Area = 18617.66 square feet

Same answer!

To solve an oblique triangle where you are given the measurement of the three sides (SSS) or two sides and the included angle (SAS), you would use the Law of Cosines:

Law of Cosines:

a

^{2}= b^{2}+ c^{2}- 2bc cos Ab

^{2}= a^{2}+ c^{2}- 2ac cos Bc

^{2}= a^{2}+ b^{2}- 2ab cos CExample 1: given SSS - find the three angles of the triangle given a = 9, b = 3 and c = 11

a

^{2}= b^{2}+ c^{2}- 2bc cos A9

^{2}= 3^{2}+ 11^{2}- 2(3)(11) cos A81 = 9 + 121 - (66) cos A

81 - 9 - 121 = - 66 cos A

-49 = -66 cos A

49/66 = Cos A

A = 42.06166467°

b

^{2}= a^{2}+ c^{2}- 2ac cos B3

^{2}= 9^{2}+ 11^{2}- 2(9)(11) cos B9 = 81 + 121 - 198 cos B

-193 = -198 cos B

193/198 = cos B

B = 12.90352054°

c

^{2}= a^{2}+ b^{2}- 2ab cos C11

^{2}= 9^{2}+ 3^{2}- 2(9)(3) cos C121 = 81 + 9 - 54 cos C

31 = -54 cos C

-31/54 = cos C

C = 125.0348148°

Example 2: Two Sides and the Included Angle - SAS

C = 108° , a = 10, b = 6.5

c

^{2}= a^{2}+ b^{2}- 2ab cos Cc

^{2}= 10^{2}+ 6.5^{2}- 2(10)(6.5) cos 108°c

^{2}= 182.4222093c = 13.50637662

Now find the other 2 angles:

a

^{2}= b^{2}+ c^{2}- 2bc cos A10

^{2}= 6.5^{2}+ 13.50637662^{2}- 2(6.5)(13.50637662) cos A100 = 224.64722093 - 175.582896 cos A

-124.6722093 = - 175.582896 cos A

.7100475736 = cos A

A = 44.76121382°

B = 180° - 108° - 44.76121382° = 27.23878618°

Example 3: If a street light is 2 feet in length on the post, the top is 3 feet and the distance from the bulb to the post is 4.5 feet. What is the angle measurement from the post to the top of the sign?

4.5

^{2}= 2^{2}+ 3^{2}- 2 (2)(3) Cos A20.25 = 4 + 9 - 12 Cos A

7.25 = -12 Cos A

-7.25/12 = Cos A

A = 127.1688997°

so therefore, the angle from the post to the top of the sign would be

180° - 127.1688997° = 52.83110034°

Heron’s Area Formula - given any triangle with sides of lengths a, b, and c, the area of the triangle is:

Given the lengths of the sides a, b, and c and the semiperimeter s,

Example 4: The Landau Building in Cambridge, Massachusetts, is a building with a triangular-shaped base. The lengths of the sides of the triangular base are 145 feet, 257 feet, and 290 feet. Find the area of the base of the building.

s = (145 + 257 + 290)/2 = 692/2 = 346 feet

Area = √(346(346 - 145)(346 - 257)(346 - 290))

= 18617.66 square feet

OR

Using Area = ½ bc Sin A, first find the measure of angle A

a

^{2}= b^{2}+ c^{2}- 2bc sin A145

^{2}= 257^{2}+ 290^{2}- 2(257)(290) cos A21025 = 66049 + 84100 - 149060 Cos A

-129124 = - 149060 cos A

129124/149060 = cos A

A = 29.97365689°

Area = ½ (257)(290) sin 29.97365689°

Area = 18617.66 square feet

Same answer!