Download The LAW of COSINES

Trigonometry
Sec. 04 notes
MathHands.com
Márquez
The LAW of COSINES
Main Idea
At the beginning of the course, we stated the essential concert of trigonometry to be the study and solving of
triangles. The general goal is to determine all 3 sides and all 3 angles of a triangles, when only 3 of these are
known. Moreover, in the case of right triangles, we have been there, done that. In deed, we promised to return to
this question and resolve not just right triangles, but any and all solvable triangles.
We are now ready to take on such task. The key ingredient will be The Law of Cosines. One may find it helpful to
think of the law of cosines as a generalization of the Pythagoras Theorem. Let us recall, the Pythagoras Theorem.
If we stand next to the 90◦ angle of a right triangle and look across, there we will find the hypothenuse, thus we
are reminded of:
Pythagoras Theorem:
A
c
B
b
a
c2 = a 2 + b 2
The question of interest now is, what if this is not a right triangle?
A
b
c
B
a
C
c2
c
2007-2009
MathHands.com
math
hands
? 2 2
a +b
=
pg. 1
Trigonometry
Sec. 04 notes
MathHands.com
Márquez
The Law of Cosines is taylor-made for exactly this purpose, to address such triangles. It states how to ’tweak’ the
pythagoras theorem to apply it to non-right triangles. Without further delay, here is the law of cosines:
Law of Cosines:
A
b
c
B
C
a
c2 = a2 + b2 −2ab cos C
In fact, the law of cosines can be applied from any corner of the triangle: observe:
a2 = c2 + b2 −2cb cos A
A
A
b
c
B
C
a
b
c
B
a
C
b2 = c2 + a2 −2ca cos B
It can even be applied to right triangles:
A
c
B
b
a
c2 = a2 + b2 −2ab cos 90◦
Notice, cos 90◦ = 0, thus in this case, the law of cosines reduces to just pythagoras.
c
2007-2009
MathHands.com
math
hands
pg. 2
Trigonometry
Sec. 04 notes
MathHands.com
Márquez
Now, that we have seen the law of cosines, there are two important tasks at hand. One, we would like to prove the
law of cosines, and two, we would like to learn how to use it to solve triangles. Let us take the latter first.
Count The Unknown Variables:
Law of cosines always involves 4 quantities from the triangle. For example, in this case,
A
b
c
B
C◦
a
c2 = a2 + b2 −2(a)b cos C ◦
the four quantities involved are c, a, b, and angle C. The use of the law of cosines would be most successful, when out
of these 4 quantities, 3 are known, since this would lead to an equation with just one unknown variable. Consider
the following example:
A
b
12
B
70◦
(10)
122 = (10)2 + b2 −2((10))b cos 70◦
In this case, we could simplify the equation, estimate cos 70◦ and solve the quadratic equation using the quadratic
formula, for example. Ultimately, we could solve for b...
However consider what would happen if we applied the law of cosines from a different corner.
(10)2 = (12)2 + b2 −2(12)b cos A◦
A
)
(12
B
b
(10)
70◦
Note, in this case, the law of cosines leads to an equation with two unknown quantities. Now, consider the choices:
solving an equation with one variable OR solving an equation with two variables. If you prefer equations with
less variables, you should apply the law of cosines on from the appropriate corner, and always be conscious that
applying it at different corners may lead to a different number of unknown variables to solve for.
c
2007-2009
MathHands.com
math
hands
pg. 3
Trigonometry
Sec. 04 notes
MathHands.com
Márquez
Example:
Note, in this example exactly 3 quantities are known, 8, 10 and 70◦ . Our mission is to find the other 3 missing
items, if such feat is possible. We first apply the law of cosines:
A
b
12
B
70◦
10
122 = 102 + b2 −2(10)b cos 70◦
(12)2 = (10)2 + (b)2 − 2(10)(b) cos(70◦ )
2
144 = 100 + (b) − 20b cos(70 )
2
◦
0 = (b) − [20 cos(70 )]b + −44
◦
OR
(BI)
(BI)
2
(BI)
b ≈ − 4.04
(Quadratic Formula)
0 ≈ (b) + −6.84b + −44
b ≈ 10.88
(THE LoC)
Furthermore, if we assume b is is a positive real number, then there is only one choice: namely, b ≈ 10.88, and we
have solved for one of the variables, leaving two more to go, B and A.
We now apply the law of cosines from the A angle, aware of the number of unknown variables from such, in this
case, one.
102 = 122 + (10.88)2−2(12)(10.88) cos A
A
88
10.
12
B
10
70◦
102 ≈ 122 + (10.88)2 − 2(12)(10.88) cos A
100 ≈ 144 + −261.12 cos A
−44 ≈ −261.12 cos A
−44
≈ cos A
−261.12
0.169 ≈ cos A
c
2007-2009
MathHands.com
math
hands
(LoC)
(bi)
(algebra)
(algebra)
(algebra)
pg. 4
Trigonometry
Sec. 04 notes
MathHands.com
Márquez
Solve
cos (A) ≈ 0.169
Solution:
Ak ≈ 80.27◦ + k360◦
for k ∈ Z
Ak ≈ 279.73◦ + k360◦
for k ∈ Z
OR
Now, if we assume A is an interior angle of a triangle, then it has to be more than zero but less than 180◦. Then
of all the possible choices above, only
A ≈ 80.27◦
works since it it the only one less than 180◦ and larger than 0.
Then to solve for B, we use the fact that
B + 70◦ + 80.27◦ = 180◦
thus...
B ≈ 29.73◦
... completely solving the triangle..
80.27◦
88
10.
12
29.73◦
c
2007-2009
MathHands.com
10
math
hands
70◦
pg. 5
Trigonometry
Sec. 04 exercises
MathHands.com
Márquez
The LAW of COSINES
1. Solve
A
b
13
B
60◦
10
2. Solve
A
b
9
B
40◦
12
3. Solve
A
13
B
b
50◦
7
4. Solve
A
b
13
B
c
2007-2009
MathHands.com
11
math
hands
65◦
pg. 6
Trigonometry
Sec. 04 exercises
MathHands.com
Márquez
5. Solve
C
3
a
20◦
B
5
6. Solve
C
5
a
35◦
B
4
7. Solve
C
b
5
A
35◦
4
8. Solve
C
b
2
A
67◦
4
9. Solve
C
3
a
40◦
4.5
c
2007-2009
MathHands.com
B
math
hands
pg. 7
Trigonometry
Sec. 04 exercises
MathHands.com
Márquez
10. Solve
C
3
a
15◦
B
1
11. Solve
C
b
2
A
50◦
3
12. PROVE the Law of COSINES
A
b
c
B
C
a
c2 = a2 + b2 −2ab cos C
c
2007-2009
MathHands.com
math
hands
pg. 8