cos B =
2 ca
42 + 22 − 32
⇒
cos B =
= 0.6875
2(4)(2)
⇒
B = 46.6◦
Now we use the formula for c 2 to find C:
a 2 + b 2 − c 2
c 2 = a 2 + b 2 − 2 ab cos C
⇒
cos C =
2 ab
22 + 32 − 42
⇒
cos C =
= −0.25
2(2)(3)
⇒
C = 104.5◦
Thus, A = 180◦ − B − C = 180◦ − 46.6◦ − 104.5◦ ⇒ A = 28.9◦ .
It may seem that there is always a solution in Case 4 (given all three sides), but that is
not true, as the following example shows.
Example 2.7
Case 4: Three sides.
C
Solve the triangle △ ABC given a = 2, b = 3, and c = 6.
b = 3
a = 2
Solution: If we blindly try to use the Law of Cosines to find A, we get
A
B
c
b 2
32
= 6
+ c 2 − a 2
+ 62 − 22
a 2 = b 2 + c 2 − 2 bc cos A
⇒
cos A =
=
= 1.139 ,
2 bc
2(3)(6)
which is impossible since |cos A| ≤ 1. Thus, there is no solution .
We could have saved ourselves some effort by recognizing that the length
b = 3
a = 2
of one of the sides ( c = 6) is greater than the sums of the lengths of the
remaining sides ( a = 2 and b = 3), which (as the picture on the right shows)
c = 6
is impossible in a triangle.
The Law of Cosines can also be used to solve triangles in Case 2 (two sides and one op-
posite angle), though it is less commonly used for that purpose than the Law of Sines. The
following example gives an idea of how to do this.
48
Chapter 2 • General Triangles
§2.2
Example 2.8
Case 2: Two sides and one opposite angle.
Solve the triangle △ ABC given a = 18, A = 25◦, and b = 30.
Solution: In Example 2.2 from Section 2.1 we used the Law of Sines to show that there are two sets
of solutions for this triangle: B = 44.8◦, C = 110.2◦, c = 40 and B = 135.2◦, C = 19.8◦, c = 14.4. To solve this using the Law of Cosines, first find c by using the formula for a 2:
a 2 = b 2 + c 2 − 2 bc cos A
⇒
182 = 302 + c 2 − 2(30) c cos 25◦
⇒
c 2 − 54.38 c + 576 = 0 ,
which is a quadratic equation in c, so we know that it can have either zero, one, or two real roots
(corresponding to the number of solutions in Case 2). By the quadratic formula, we have
54.38 ±
(54.38)2 − 4(1)(576)
c =
= 40 or 14.4 .
2(1)
Note that these are the same values for c that we found before. For c = 40 we get
c 2 + a 2 − b 2
402 + 182 − 302
cos B =
=
= 0.7111
⇒
B = 44.7◦
⇒
C = 110.3◦ ,
2 ca
2(40)(18)
which is close to what we found before (the small difference being due to different rounding). The
other solution set can be obtained similarly.
Like the Law of Sines, the Law of Cosines can be used to prove some geometric facts, as
in the following example.
Example 2.9
Use the Law of Cosines to prove that the sum of the squares of the
D
b
diagonals of any parallelogram equals the sum of the squares of the
sides.
c
a
a
Solution: Let a and b be the lengths of the sides, and let the diago-
nals opposite the angles C and D have lengths c and d, respectively,
d
as in Figure 2.2.2. Then we need to show that
C
b
c 2 + d 2 = a 2 + b 2 + a 2 + b 2 = 2( a 2 + b 2) .
Figure 2.2.2
By the Law of Cosines, we know that
c 2 = a 2 + b 2 − 2 ab cos C , and
d 2 = a 2 + b 2 − 2 ab cos D .
By properties of parallelograms, we know that D = 180◦ − C, so
d 2 = a 2 + b 2 − 2 ab cos (180◦ − C)
= a 2 + b 2 + 2 ab cos C ,
since cos (180◦ − C) = −cos C. Thus,
c 2 + d 2 = a 2 + b 2 − 2 ab cos C + a 2 + b 2 + 2 ab cos C
= 2( a 2 + b 2) .
QED
The Law of Cosines • Section 2.2
49
Exercises
For Exercises 1-6, solve the triangle △ ABC.
1. A = 60◦, b = 8, c = 12
2. A = 30◦, b = 4, c = 6
3. a = 7, B = 60◦, c = 9
4. a = 7, b = 3, c = 9
5. a = 6, b = 4, c = 1
6. a = 11, b = 13, c = 16
7. The diagonals of a parallelogram intersect at a 42◦ angle and have lengths of 12 and 7 cm. Find
the lengths of the sides of the parallelogram. ( Hint: The diagonals bisect each other. )
8. Two trains leave the same train station at the same time, moving along straight tracks that form
a 35◦ angle. If one train travels at an average speed of 100 mi/hr and the other at an average
speed of 90 mi/hr, how far apart are the trains after half an hour?
9. Three circles with radii of 4, 5, and 6 cm, respectively, are tangent to each other externally. Find
the angles of the triangle whose vertexes are the centers of the circles.
10. Find the length x of the diagonal of the quadrilateral in Figure 2.2.3 below.
4
3.5
60◦
x
5
3
6
2
5.5
Figure 2.2.3
Exercise 10
Figure 2.2.4
Exercise 11
11. Two circles of radii 5 and 3 cm, respectively, intersect at two points. At either point of intersec-
tion, the tangent lines to the circles form a 60◦ angle, as in Figure 2.2.4 above. Find the distance
between the centers of the circles.
12. Use the Law of Cosines to show that for any triangle △ ABC, c 2 < a 2+ b 2 if C is acute, c 2 > a 2+ b 2
if C is obtuse, and c 2 = a 2 + b 2 if C is a right angle.
13. Show that for any triangle △ ABC,
cos A
cos B
cos C
a 2 + b 2 + c 2
+
+
=
.
a
b
c
2 abc
14. Show that for any triangle △ ABC,
a 2 ( b + c − a) + b 2 ( a + c − b) + c 2 ( a + b − c) cos A + cos B + cos C =
.
2 abc
What do the terms in parentheses represent geometrically? Use your answer to explain why
cos A + cos B + cos C > 0 for any triangle, even if one of the cosines is negative.2
15. Prove the Law of Cosines (i.e. formulas (2.9)-(2.11)) for right triangles.
16. Recall from elementary geometry that a median of a triangle is a line segment from any vertex
to the midpoint of the opposite side. Show that the sum of the squares of the three medians of a
triangle is 3⁄4 the sum of the squares of the sides.
2It turns out that 1 < cos A + cos B + cos C ≤ 3/2 for any triangle, as we will see later.
50
Chapter 2 • General Triangles
§2.2
17. The Dutch astronomer and mathematician Willebrord Snell (1580-1626) wrote the Law of Cosines
as
2 ab
1
=
c 2 − ( a − b)2
1 − cos C
in his trigonometry text Doctrina triangulorum (published a year after his death). Show that this
formula is equivalent to formula (2.11) in our statement of the Law of Cosines.
18. Suppose that a satellite in space, an earth station, and the center of the earth all lie in the same
plane. Let re be the radius of the earth, let rs be the distance from the center of the earth to the
satellite (called the orbital radius of the satellite), and let d be the distance from the earth station
to the satellite. Let E be the angle of elevation from the earth station to the satellite, and let γ
and ψ be the angles shown in Figure 2.2.5.
satellite
✸
local horizontal
d
rs
E
ψ
γ
❚
r
earth station
e
center
of earth
Figure 2.2.5
Use the Law of Cosines to show that
r
2
e
r e
d = rs
1 +
− 2
cos γ ,
rs
rs
and then use E = ψ − 90◦ and the Law of Sines to show that
sin γ
cos E =
.
r
2
e
r e
1 +
− 2
cos γ
rs
rs
Note: This formula allows the angle of elevation E to be calculated from the coordinates of the
earth station and the subsatellite point (where the line from the satellite to the center of the earth
crosses the surface of the earth).3
3See pp. 22-25 in T. PRATT AND C.W. BOSTIAN, Satellite Communications, New York: John Wiley & Sons,
1986.
The Law of Tangents • Section 2.3
51
2.3 The Law of Tangents
We have shown how to solve a triangle in all four cases discussed at the beginning of this
chapter. An alternative to the Law of Cosines for Case 3 (two sides and the included angle)
is the Law of Tangents:
Theorem 2.3. Law of Tangents: If a triangle has sides of lengths a, b, and c opposite the
angles A, B, and C, respectively, then
a − b
tan 1 ( A − B)
=
2
,
(2.17)
a + b
tan 1 ( A
2
+ B)
b − c
tan 1 ( B − C)
=
2
,
(2.18)
b + c
tan 1 ( B
2
+ C)
c − a
tan 1 ( C − A)
=
2
.
(2.19)
c + a
tan 1 ( C
2
+ A)
Note that since tan (− θ) = −tan θ for any angle θ, we can switch the order of the letters in
each of the above formulas. For example, we can rewrite formula (2.17) as
b − a
tan 1 ( B − A)
=
2
,
(2.20)
b + a
tan 1 ( B
2
+ A)
and similarly for the other formulas. If a > b, then it is usually more convenient to use
formula (2.17), while formula (2.20) is more convenient when b > a.
Example 2.10
Case 3: Two sides and the included angle.
C = 96◦
Solve the triangle △ ABC given a = 5, b = 3, and C = 96◦.
b = 3
a = 5
Solution: A + B + C = 180◦, so A + B = 180◦ − C = 180◦ − 96◦ = 84◦. Thus, by
the Law of Tangents,
c
A
B
a − b
tan 1 ( A − B)
5 − 3
tan 1 ( A − B)
=
2
⇒
=
2
a + b
tan 1 ( A
5 + 3
tan 1 (84◦)
2
+ B)
2
⇒
tan 1 ( A
tan 42◦
2
− B) = 28
= 0.2251
⇒
1 ( A
2
− B) = 12.7◦
⇒
A − B = 25.4◦ .
We now have two equations involving A and B, which we can solve by adding the equations:
A − B = 25.4◦
A + B = 84◦
−− − − − − − −
2 A
= 109.4◦
⇒
A = 54.7◦
⇒
B = 84◦ − 54.7◦
⇒
B = 29.3◦
We can find the remaining side c by using the Law of Sines:
a sin C
5 sin 96◦
c =
=
⇒
c = 6.09
sin A
sin 54.7◦
52
Chapter 2 • General Triangles
§2.3
Note that in any triangle △ ABC, if a = b then A = B (why?), and so both sides of formula
(2.17) would be 0 (since tan 0◦ = 0). This means that the Law of Tangents is of no help in
Case 3 when the two known sides are equal. For this reason, and perhaps also because of the
somewhat unusual way in which it is used, the Law of Tangents seems to have fallen out of
favor in trigonometry books lately. It does not seem to have any advantages over the Law
of Cosines, which works even when the sides are equal, requires slightly fewer steps, and is
perhaps more straightforward.4
Related to the Law of Tangents are Mollweide’s equations:5
Mollweide’s equations: For any triangle △ ABC,
a − b
sin 1 ( A − B)
=
2
, and
(2.21)
c
cos 1 C
2
a + b
cos 1 ( A − B)
=
2
.
(2.22)
c
sin 1 C
2
Note that all six parts of a triangle appear in both of Mollweide’s equations. For this
reason, either equation can be used to check a solution of a triangle. If both sides of the
equation agree (more or less), then we know that the solution is correct.
Example 2.11
Use one of Mollweide’s equations to check the solution of the triangle from Example 2.10.
Solution: Recall that the full solution was a = 5, b = 3, c = 6.09, A = 54.7◦, B = 29.3◦, and C = 96◦.
We will check this with equation (2.21):
a − b
sin 1 ( A − B)
=
2
c
cos 1 C
2
5 − 3
sin 1 (54.7◦ − 29.3◦)
=
2
6.09
cos 1 (96◦)
2
2
sin 12.7◦
=
6.09
cos 48◦
0.3284 = 0.3285 ✓
The small difference (≈ 0.0001) is due to rounding errors from the original solution, so we can con-
clude that both sides of the equation agree, and hence the solution is correct.
4Before the advent of electronic calculators, the Law of Tangents was more popular than it is today since
it lent itself better than the Law of Cosines to what was known as logarithmic computation. In those days,
computations with large numbers were handled by taking logarithms and looking up values in a logarithm table.
Ratios (such as in the Law of Tangents and the Law of Sines) could be replaced by differences of logarithms,
making computation easier.
5Named