Download 411 Exercise 2 Law of Sines 15–3 Law of Cosines

Section 15–3
Exercise 2
◆
◆
411
Law of Cosines
Law of Sines
Data for triangle ABC are given in the following table. Solve for the missing parts.
Angles
A
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
Sides
B
C
a
46.3
228
65.9
43.0
35.17
21.45
61.9
126
b
304
c
1.59
1.46
15.0
21.0
276.0
47.0
7.65
27.0
119
31.6
44.8
11.7
72.0
15.0
125
375
32.0
58.0
38.27
24.14
5562
55.38
18.20
77.85
63.88
44.47
18.0
12.0
1.065
50.7
45.55
1137
1010
15–3 Law of Cosines
Derivation
Consider an oblique triangle ABC as shown in Fig. 15–17. As we did for the law of sines, we
start by dividing the triangle into two right triangles by drawing an altitude h to side AC.
In right triangle ABD,
c2 h2 (AD)2
But AD b CD. Substituting, we get
c2 h2 (b CD)2
Now, in right triangle BCD, by Eq. 147,
CD
a
cos C
or
CD a cos C
B
c
a
h
A
C
D
b
FIGURE 15–17 Derivation of the
law of cosines.
Substituting a cos C for CD in Eq. (1) yields
c2 h2 (b a cos C)2
(1)
412
Chapter 15
◆
Oblique Triangles and Vectors
Squaring, we have
c2 h2 b2 2ab cos C a2 cos2 C
(2)
Let us leave this expression for the moment and write the Pythagorean theorem for the
same triangle BCD.
h2 a2 (CD)2
Again substituting a cos C for CD, we obtain
h2 a2 (a cos C)2
a2 a2 cos2 C
Substituting this expression for h2 back into (2), we get
c2 a2 a2 cos2 C b2 2ab cos C a2 cos2 C
Collecting terms, we get the law of cosines.
c2 a2 b2 2ab cos C
Most students find it easier to
think of the law of cosines
in terms of the parts of the
triangle, that is, the given angle,
the side opposite to the given
angle, and the sides adjacent to
the given angle, rather than by
letters of the alphabet.
Notice that when the angle in
these equations is 90, the
law of cosines reduces to the
Pythagorean theorem.
We can repeat the derivation, with perpendiculars drawn to side AB and to side BC, and get
two more forms of the law of cosines.
a2 b2 c2 2bc cos A
b2 a2 c2 2ac cos B
c2 a2 b2 2ab cos C
Law of
Cosines
141
The square of any side equals the sum of the squares of the other two sides minus twice the
product of the other sides and the cosine of the opposite angle.
◆◆◆
Example 21: Find side x in Fig. 15–18.
24
m
x
1.
42.8˚
1.87 m
FIGURE 15–18
Solution: By the law of cosines,
x2 (1.24)2 (1.87)2 2(1.24)(1.87) cos 42.8
5.03 4.64(0.7337) 1.63
x 1.28 m
◆◆◆
When to Use the Law of Sines or the Law of Cosines
It is sometimes not clear whether to use the law of sines or the law of cosines to solve a triangle.
We use the law of sines when we have a known side opposite a known angle. We use the law of
cosines only when the law of sines does not work, that is, for all other cases. In Fig. 15–19, the
heavy lines indicate the known information and might help in choosing the proper law.
Section 15–3
◆
413
Law of Cosines
AAS
Find angle first,
using Eq. 139
May have two
solutions
SSA
ASA
Use law of sines
SAS
SSS
Use law of cosines
FIGURE 15–19
When to use the law of sines or the law of cosines.
Using the Cosine Law When Two Sides
and the Included Angle Are Known
We can solve triangles by the law of cosines if we know two sides and the angle between them,
or if we know three sides. We consider the first of these cases in the following example.
◆◆◆
Example 22: Solve triangle ABC where a 184, b 125, and C 27.2.
B
Solution: We make a sketch as shown in Fig. 15–20.
Then, by the law of cosines,
c a b 2ab cos C
(184)2 (125)2 2(184)(125) cos 27.2
2
2
2
a = 184
c
27.2˚
c 兹 (184)2 (125)2 2(184)(125) cos 27.2
92.6
A
b = 125
FIGURE 15–20
Now that we have a known side opposite a known angle, we can use the law of sines to find
angle A or angle B. Which shall we find first?
Use the law of sines to find the acute angle first (angle B in this example). If, instead, you
solve for the obtuse angle first, you might forget to subtract the angle obtained from the calculator from 180. Further, if one of the angles is so close to 90 that you cannot tell from your
sketch if it is acute or obtuse, find the other angle first, and then subtract the two known angles
from 180 to obtain the third angle.
By the law of sines,
sin B
125
sin 27.2
92.6
125 sin 27.2
sin B 0.617
92.6
B 38.1 and B 180 38.1 141.9
We drop the larger value because our sketch shows us that B must be acute. Then, by Eq. 139,
A 180 27.2 38.1 114.7
◆◆◆
Notice that we cannot initially
use the law of sines because
we do not have a known side
opposite a known angle.
C
414
Chapter 15
◆
Oblique Triangles and Vectors
In our next example, the given angle is obtuse.
B
◆◆◆
Example 23: Solve triangle ABC where b 16.4, c 10.6, and A 128.5.
a
.6
10
Solution: We make a sketch as shown in Fig. 15–21. Then, by the law of cosines,
128.5˚
A
a2 (16.4)2 (10.6)2 2(16.4)(10.6) cos 128.5
C
16.4
FIGURE 15–21
Common
Error
The cosine of an obtuse angle is negative. Be sure to use the
proper algebraic sign when applying the law of cosines to an
obtuse angle.
In our example,
cos 128.5 0.6225
So
a 兹(16.4)2 (10.6)2 2(16.4)(10.6)(0.6225) 24.4
By the law of sines,
sin B
16.4
sin 128.5
24.4
16.4 sin 128.5
sin B 0.526
24.4
B 31.7 and B 180 31.7 148.3
We drop the larger value because our sketch shows us that B must be acute. Then, by Eq. 139,
C 180 31.7 128.5 19.8
◆◆◆
Using the Cosine Law When Three Sides Are Known
When three sides of an oblique triangle are known, we can use the law of cosines to solve for
one of the angles. A second angle is found using the law of sines, and the third angle is found
by subtracting the other two from 180.
Example 24: Solve triangle ABC in Fig. 15–22, where a 128, b 146, and
c 222.
◆◆◆
C
146
8
12
Solution: We start by writing the law of cosines for any of the three angles. A good way
to avoid ambiguity is to find the largest angle first (the law of cosines will tell us if it is
acute or obtuse), and then we are sure that the other two angles are acute.
Writing the law of cosines for angle C gives
A
222
B
FIGURE 15–22
(222)2 (128)2 (146)2 2(128)(146) cos C
Solving for cos C gives
cos C 0.3099
Since the cosine is negative, C must be obtuse, so
C 108.1
◆
Section 15–4
415
Applications
Then by the law of sines
sin A
128
sin 108.1
222
from which sin A 0.5480. Since we know that A is acute, we get
A 33.2
Finally,
B 180 108.1 33.2 38.7
Exercise 3
◆
◆◆◆
Law of Cosines
Data for triangle ABC are given in the following table. Solve for the missing parts.
Angles
A
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B
Sides
C
a
b
106.0
15.7
11.2
51.4
1.95
128
728
128
1.16
68.3
27.3
135
35.2
41.7
77.3
11.3
199
129
1.475
9.08
186
115
67.0
158
41.77
108.8
36.29
41.8
1445
7.286
97.3
18.3
906
152
275
81.4
15.6
46.8
1.836
6.75
1.77
57.2
1502
81.4
47.28
c
1.46
21.7
663
1.95
214
12.8
202
51.3
2.017
179
1.99
36.7
6.187
88.5
51.36
15–4 Applications
As with right triangles, oblique triangles have many applications in technology, as you will see
in the exercises for this section. Follow the same procedures for setting up these problems as we
used for other word problems, and solve the resulting triangles by the law of sines or the law
of cosines, or both.
If an area of an oblique triangle is needed, either compute all the sides and use Hero’s formula (Eq. 138), or find an altitude with right-triangle trigonometry and use Eq. 137.