The document discusses the Law of Cosines and how to use it to solve for unknown sides and angles of triangles given certain known information. It provides examples of using the Law of Cosines to calculate side lengths and angles of several triangles when some combination of sides and/or angles are given. It also gives an example word problem about finding the perimeter of a triangular lot needed to enclose it with chicken wire.

2.  Law of Cosines
◦ c^2 = a^2 + b^2 - ; cos C = a^2 + b^2 – c^2/2ab
◦ a^2 = b^2 + c^2 – 2bc cos A ; cos A = b^2 + c^2 – a^2/2bc
◦ b^2 = a^2 + c^2 – 2ac cos B ; cos B = a^2 + c^2 – b^2

3.  Solve for the unknown parts of ABC
A
B
C
c = 68
a = 51
b = ?
37^o

4.  Since the given sides are a & c, to solve for b we use the
formula:
 b^2 = a^2 + c^2 – 2ac cos B, substitute the given values
 = (51)^2 + (68)^2 – 2(51)(68) cos 37^o
 = 2601 + 4624 – 6936(0.7986)
 = 7225 – 5539.1
 = 1686.1
 b = 41 (Rounded to the nearest ones)

5.  To solve for A, we use formula:
 cos A = b^2 + c^2 – a^2/2bc
 = (41)^2 + (68)^2 – (51)^2/2(41)(68)
 = cos A = 0.6643
 Therefore A = 48^o22’
 To solve for C, we know that A + B + C = 180^o
 Therefore C = 180^o – (A + B)
= 180^o – (48^o22’ + 37^o)
= 94^o38’

6.  Solve for the angle of ABC
C
b = 18 a = 23
A B
c = 29

7.  To solve for A, we use the formula:
cos A = b^2 + c^2 – a^2/2bc
= (18)^2 + (29)^2 – (23)^2/2(18)(29)
= 0.06092
Therefore A = 52^o28’
 To solve for B, the formula is:
cos B = a^2 + c^2 – b^2/2ac
= (23)^2 + (29)^2 – (18)^2/2(23)(29)
Therefore B = 38^o22’
 To solve for C, the formula is:
cos C = a^2 + b^2 – c^2 /2ab
= (18)^2 + (23)^2 – (29)^2/2(23)(18)
=0.0145
Therefore C = 89^o10’
Hence A + B + C = 180

9.  Problem:
◦ Two sides of triangular lot measures 33 and 28 meters. If the sides form
an angle 55^o between them, how many meter of chicken wire be
needed to enclose the lot?
55^o
33
28

10.  To solve the problem, we have to solve for the
perimeter of the triangle, therefore we have to solve
first for the third side.
◦ Let b = to the third side:
b^2 = a^2 + c^2 – 2ac cos B; substitute the values
= (28)^2 + (33)^2 – 2(28)(33) cos 55^o
= 784 + 1089 – 1288(0.5736)
= 7873 – 739
= 7134
◦ Therefore b = 84.5 m
◦ The perimeter of the triangle is 145.5 m; therefore the
chicken wire needed to enclose the lot is 145.5 meters

11. a b c A B C
1. 12 cm 24 cm 18 cm 29° 108° 43°
2. 93.6 cm 56.5 cm 67.2 cm 98° 50’ 36.7° 45°
3. 105 cm 13 cm 85 cm 53° 87° 20’ 40°
4. 38.4 m 48.8 m 47.7 cm 46.9° 68° 65° 50’
5. 208 cm 208.5 cm 109.5 cm 74.5° 75° 40’ 30.5°
6. 23 m 19 m 27 m 56.9° 43.8° 79.3°
7. 37 m 48 m 52 m 43° 62.6° 74.2°
8. 103 cm 176 m 151 cm 35.7° 85.6° 58.8°
9. 47.7 m 36.4 m 41.5 m 75.2° 47.5° 57.3°
10. 89.6 m 108 m 94.2 m 52° 72° 56°