Download Law of Cosines

Law of Cosines
Objective: The student will identify, create,
and solve practical problems involving
triangles. Techniques will include using
the trigonometric functions, The
Pythagorean Theorem, the Law of Sines,
and the Law of Cosines.
Law of Cosines
• The Law of Sines can be used when we know
one complete pair within a triangle
• The size of an angle and the length of the side
across from that angle.
• If this is not the case, however, we need
another method for solving non-right
triangles.
• This method is called the Law of Cosines.
Law of Cosines
• Need three pieces of information in the
triangle
• Two sides and the angle between those sides
(SAS)
• All three sides (SSS)
• When you have the option, always find the
angles in order from largest to smallest if
you use the Law of Cosines, or smallest to
largest if you use the Law of Sines.
Law of Cosines
• The Law of Cosines can take three
different forms, depending upon which
sides and/or angle you are given.
• a2 = b2 + c2 – 2bc • cos A
• b2 = a2 + c2 – 2ac • cos B
• c2 = a2 + b2 – 2ab • cos C
Law of Cosines
• The third version looks a lot like the
Pythagorean Theorem. The Pythagorean
Theorem is a special case of the Law of
Cosines.
c2 = a2 + b2 – 2ab • cos C
c2 = a2 + b2 – 2ab • cos 90°
c2 = a2 + b2 – 2ab • 0
c2 = a2 + b2 – 0
c2 = a2 + b2
Note: cos 90° = 0
Law of Cosines
• If you know all three sides, and need to find
the angles, the three formulas can be
rewritten in terms of the angle as follows:
b2  c2  a 2
• cos A =
2bc
a 2  c2  b2
• cos B =
2ac
a 2  b2  c2
• cos C =
2ab
Example 1:
• Solve the triangle where a = 5, b = 7, and C = 43°
B
c
A
5
43°
7
C
c2 = a2 + b2 – 2ab • cos C
c2 = 52 + 72 – 2(5)(7) • cos 43°
sin 43 sin A

4.7754
5
4.7754 • sin A = 5 • sin 43°
5  sin 43
sin A =
4.7754
sin A = 0.71406
c2 = 22.8052
c = 4.7754
c = 4.8
B = 180 – 43 – 46
B = 91°
A = sin-1(0.71406)
A = 45.56°
A = 46°
Example 2:
• Solve the triangle where a = 12, b = 17, and c = 23
B
a2  c 2  b2
a2  b2  c 2
cos B 
cos C 
23
12
2ac
2ab
122  232  172
122  172  232
cos B 
cos C 
2 12  23 
2 12 17 
C
A
17
384
96
cos B 
cos C 
552
408
1  384 
1  96 
B

cos
C

cos
 552 
 408 
A = 180 – 104 – 46




A = 30°
B = 45.9207
C = 103.6089
B = 46°
C = 104°
Example 3:
• A road goes up a hill for 895 feet, then flattens out by 8°
and continues up the hill for another 542 feet.
Engineers want to build a road that climbs the hill at a
constant slope. What would be the length of the new
C
road?
542 ft
8°
B
895 ft
A
172°
b
b2 = a2 + c2 – 2ac • cos B
b2 = 5422 + 8952 – 2(542)(895) • cos 172°
b2 = 2,055,527.275
b = 1433.711015
b = 1434 ft