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:
a2 = b2 + c2 - 2bc cos A
b2 = a2 + c2 - 2ac cos B
c2 = a2 + b2 - 2ab cos C
Example 1: given SSS - find the three angles of the triangle given a = 9, b = 3 and c = 11
a2 = b2 + c2 - 2bc cos A
92 = 32 + 112 - 2(3)(11) cos A
81 = 9 + 121 - (66) cos A
81 - 9 - 121 = - 66 cos A
-49 = -66 cos A
49/66 = Cos A
A = 42.06166467°
b2 = a2 + c2 - 2ac cos B
32 = 92 + 112 - 2(9)(11) cos B
9 = 81 + 121 - 198 cos B
-193 = -198 cos B
193/198 = cos B
B = 12.90352054°
c2 = a2 + b2 - 2ab cos C
112 = 92 + 32 - 2(9)(3) cos C
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
c2 = a2 + b2 - 2ab cos C
c2 = 102 + 6.52 - 2(10)(6.5) cos 108°
c2 = 182.4222093
c = 13.50637662
Now find the other 2 angles:
a2 = b2 + c2 - 2bc cos A
102 = 6.52 + 13.506376622 - 2(6.5)(13.50637662) cos A
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?
4.52 = 22 + 32 - 2 (2)(3) Cos A
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
a2 = b2 + c2 - 2bc sin A
1452 = 257 2 + 290 2 - 2(257)(290) cos A
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:
a2 = b2 + c2 - 2bc cos A
b2 = a2 + c2 - 2ac cos B
c2 = a2 + b2 - 2ab cos C
Example 1: given SSS - find the three angles of the triangle given a = 9, b = 3 and c = 11
a2 = b2 + c2 - 2bc cos A
92 = 32 + 112 - 2(3)(11) cos A
81 = 9 + 121 - (66) cos A
81 - 9 - 121 = - 66 cos A
-49 = -66 cos A
49/66 = Cos A
A = 42.06166467°
b2 = a2 + c2 - 2ac cos B
32 = 92 + 112 - 2(9)(11) cos B
9 = 81 + 121 - 198 cos B
-193 = -198 cos B
193/198 = cos B
B = 12.90352054°
c2 = a2 + b2 - 2ab cos C
112 = 92 + 32 - 2(9)(3) cos C
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
c2 = a2 + b2 - 2ab cos C
c2 = 102 + 6.52 - 2(10)(6.5) cos 108°
c2 = 182.4222093
c = 13.50637662
Now find the other 2 angles:
a2 = b2 + c2 - 2bc cos A
102 = 6.52 + 13.506376622 - 2(6.5)(13.50637662) cos A
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?
4.52 = 22 + 32 - 2 (2)(3) Cos A
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:
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
a2 = b2 + c2 - 2bc sin A
1452 = 257 2 + 290 2 - 2(257)(290) cos A
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!
