Download - 6.2 Law of Cosines
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