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In this chapter a number of additional topics involving trigonometry are
considered. First, the problem of solving triangles is returned to—not just
right triangles but any triangle. Then some of these ideas are used to
develop the important concept of a vector. With knowledge of trigonometry, the polar coordinate system is introduced, probably the most important coordinate system after the rectangular coordinate system. After considering polar equations and their graphs, complex numbers are
represented in polar form. Once a complex number is in polar form, it
will be possible to find nth powers and nth roots of the number using an
ingenious theorem due to De Moivre.
SECTION
7-1
Law of Sines
• Law of Sines Derivation
• Solving the ASA and AAS Cases
• Solving the SSA Case—Including the Ambiguous Case
The law of sines (developed in this section) and the law of cosines (developed in the
next section) play fundamental roles in solving oblique triangles—triangles without
a right angle. Every oblique triangle is either acute, all angles between 0° and 90°, or
obtuse, one angle between 90° and 180°. Figure 1 illustrates both types of triangles.
FIGURE 1 Oblique triangles.
b
a
c
a
b
c
Acute triangle
Obtuse triangle
(a)
TABLE 1
Triangles and
Significant Digits
Angle to
nearest
1°
10 or 0.1°
Significant
digits for side
measure
2
3
1 or 0.01°
4
10 or 0.001°
5
(b)
Note how the sides and angles of the oblique triangles in Figure 1 have been
labeled: Side a is opposite angle , side b is opposite angle , and side c is opposite
angle . Also note that the largest side of a triangle is opposite the largest angle. Given
any three of the six quantities indicated in Figure 1, we are interested in finding the
remaining three, if they exist. This process is called solving the triangle.
The law of sines is developed in this section, and the law of cosines is developed
in the next section. These two laws provide the basic tools for solving oblique triangles. If the given quantities include a side and the opposite angle, the law of sines
Before proceeding with specific examples, it is important to recall the rules in
Table 1 regarding accuracy of angle and side measure. Table 1 is also repeated inside
the back cover of the text for easy reference.
7-1
Law of Sines
507
Calculator Calculations
When solving for a particular side or angle, carry out all operations within the
calculator and then round to the appropriate number of significant digits (as
specified in Table 1) at the end of the calculation. Your answer may still differ
slightly from those in the book, depending on the order in which you solve for
the sides and angles.
• Law of Sines
Derivation
The law of sines is relatively easy to prove using the right triangle properties studied
in Section 5-5. We will also use the fact that
sin (180° x) sin x
which is readily obtained using a difference identity (a good exercise for you). Referring to the triangles in Figure 2, we proceed as follows: For each triangle,
b
a
h
sin h
b
h
a
and
sin and
h a sin m
Solving each equation for h, we obtain
c
h b sin (a)
Thus,
180 b
a
h
b sin a sin m
sin sin a
b
c
(b)
FIGURE 2
(1)
Similarly, for each triangle in Figure 2,
sin m
c
sin sin (180° ) and
m
a
Solving each equation for m, we obtain
m c sin and
m a sin Thus,
c sin a sin sin sin a
c
If we combine equations (1) and (2), we obtain the law of sines.
(2)
508
Theorem 1
Law of Sines
b
sin sin sin a
b
c
a
c
In words, the ratio of the sine of an angle to its opposite side is the same as the
ratio of the sine of either of the other angles to its opposite side.
Suppose that an angle of a triangle and its opposite side are known. Then the
ratio of Theorem 1 can be calculated. So if one additional part of the triangle, either
of the other angles or either of the other sides, is known, then the law of sines can
be used to solve the triangle.
Thus, the law of sines is used to solve triangles, given:
1. Two sides and an angle opposite one of them (SSA)
2. Two angles and any side (ASA or AAS)
If the given information for a triangle consists of two sides and the included angle
(SAS) or three sides (SSS), then the law of sines cannot be applied. The key to handling these two cases, the law of cosines, is developed in Section 7-2.
We will apply the law of sines to the easier ASA and AAS cases first, then we
will turn to the more challenging SSA case.
• Solving the ASA
and AAS Cases
EXAMPLE 1
Solving the ASA Case
Solve the triangle in Figure 3.
FIGURE 3
b
280
a
4520
120 meters
Solution
We are given two angles and the included side, which is the ASA case. Find the third
angle, then solve for the other two sides using the law of sines.
7-1
Law of Sines
509
180°
Solve for 180° ( )
180° (28°0 45°20)
106°40
sin sin a
c
Solve for a
a
c sin sin 120 sin 28°0
sin 106°40
58.8 meters
sin sin b
c
Solve for b
b
c sin sin 120 sin 45°20
sin 106°40
89.1 meters
Matched Problem 1
FIGURE 4
Solve the triangle in Figure 4.
b
13 0
c
35
6520
Note that the AAS case can always be converted to the ASA case by first solving for the third angle. For the ASA or AAS case to determine a unique triangle, the
sum of the two angles must be between 0° and 180°, since the sum of all three angles
in a triangle is 180° and no angle can be zero or negative.
• Solving the SSA
Case—Including the
Ambiguous Case
We now look at the case where we are given two sides and an angle opposite
one of the sides–the SSA case. This case has several possible outcomes, depending on the measures of the two sides and the angle. Table 2 illustrates the various
possibilities.
510
TABLE 2
SSA Variations
a
(h b sin )
Acute
0ah
Acute
Acute
Number of
triangles
Figure
0
ah
1
hab
2
Acute
ab
1
Obtuse
0ab
0
Case
h
b
h a
(b)
b
a
ab
h
b
a
Ambiguous
case
(c)
a
(d)
(e)
a
b
Obtuse
(a)
a
b
1
b
a
(f )
Table 2 need not be committed to memory. Usually a rough sketch of a particular situation will indicate which of the variations applies. The case where h a b
is referred to as the ambiguous case, because two triangles, one acute and the other
obtuse, are always possible.
EXPLORE-DISCUSS 1
Discuss which cases in Table 2 apply and why if in the solution process of solving an SSA triangle with acute it is found that:
(1) sin 1
(2) sin 1
(3) 0 sin 1
EXAMPLE 2
Solving the SSA Case
Solve the triangle(s) with 123°, b 23 centimeters, and a 47 centimeters.
7-1
Solution
511
Law of Sines
From a rough sketch (Fig. 5), we see that there is only one triangle:
FIGURE 5
23 cm
47 cm
123
c
sin sin b
a
Solve for b sin 23 sin 123°
a
47
23 sin 123°
sin1
24°
47
sin 180°
180° 123° 24° 33°
Solve for sin sin a
c
a sin 47 sin 33°
c
31 centimeters
sin sin 123°
Solve for c
Matched Problem 2
EXAMPLE 3
Solve the triangle(s) with 98°, a 62 meters, and b 88 meters.
Solving the SSA (Ambiguous) Case
Solve the triangle(s) with 26°, a 1.0 meter, and b 1.8 meters.
Solution
FIGURE 6
If we try to draw a triangle with the indicated sides and angle, we find that two triangles, I and II, are possible (Fig. 6). This is verified by the fact that h a b,
where h b sin 0.79 meters, a 1.0 meter, and b 1.8 meters.
I
D
1.8 m
1.0 m
A
26
B
Solve for and C
A
26
c
D
1.8 m
1.8 m
1.0 m
II
D
1.0 m
B
A
26
We start by finding and using the law of sines:
sin sin b
a
b sin 1.8 sin 26°
sin 0.7891
a
1.0
1.0 m
c
C
512
Angle can be either obtuse or acute:
180° sin1 0.7891
sin1 0.7891
or
180° 52° 128°
Solve for and 52°
We next find and :
180° (26° 128°) 26°
180° (26° 52°) 102°
Solve for c and c
Finally, we solve for c and c:
sin sin a
c
c
a sin sin 1.0 sin 26°
sin 26°
1.0 meter
sin sin a
c
c a sin sin 1.0 sin 102°
sin 26°
2.2 meters
In summary:
Triangle I:
Triangle II:
Matched Problem 3
128°
52°
26°
c 1.0 meter
102°
c 2.2 meters
Solve the triangle(s) with a 8 kilometers, b 10 kilometers, and 35°.
The law of sines is useful in many applications, as can be seen in Example 4 and
the applications in Exercise 7-1.
EXAMPLE 4
Surveying
To measure the length d of a lake (see Fig. 7), a baseline AB is established and measured to be 125 meters. Angles A and B are measured to be 41.6° and 124.3°, respectively. How long is the lake?
7-1
Law of Sines
513
FIGURE 7
A
B
e
in rs
l
e
se et 124.3
Ba m
5
2
1
41.6
d
C
Find angle C and use the law of sines.
Solution
Angle C 180° (124.3° 41.6°)
14.1°
sin 14.1° sin 41.6°
125
d
d 125
41.6°
sin
sin 14.1° 341 meters
Matched Problem 4
In Example 4, find the distance AC.
1. 101°40, b 141, c 152
2. 44°, 38°, c 55 m
3. 134°, 46°, 11°, 99°, c 2.7 km, c 14 km
4. 424 m
EXERCISE
7-1
The labeling in the figure below is the convention we will
may differ slightly from those in the book, depending on the
order in which you solve for the sides and angles of a given
triangle.
b
a
c
A
Solve each triangle in Problems 1–8.
1. 73°, 28°, c 42 feet
2. 41°, 33°, c 21 centimeters
3. 122°, 18°, b 12 kilometers
4. 43°, 36°, a 92 millimeters
5. 112°, 19°, c 23 yards
6. 52°, 105°, c 47 meters
7. 52°, 47°, a 13 centimeters
8. 83°, 77°, c 25 miles
In Problems 9–20, are there zero, one, or two triangles with
the given measurements? (Do not solve the triangle.) Which
case, if any, of Table 2 applies? Explain.
9. a 5 inches, b 6 inches, 30°
10. a 6 inches, b 5 inches, 30°
11. a 4 feet, b 8 feet, 30°
514
12. a 2 feet, b 5 feet, 30°
an error has been made in solving a triangle. Use this equation to check Problem 1. (Because of rounding errors, both
sides may not be exactly the same.)
13. a 5 inches, b 8 inches, 45°
14. a 3 inches, b 4 inches, 45°
36. (A) Use the law of sines and suitable identities to show that
for any triangle
15. a 4 feet, b 5 feet, 120°
16. a 6 feet, b 4 feet, 120°
ab
a
b
17. a 8 inches, b 6 inches, 60°
18. a 7 inches, b 8 inches, 60°
19. a 13 feet, b 12 feet, 90°
2
tan
2
tan
(B) Verify the formula with values from Problem 1.
20. a 12 feet, b 13 feet, 90°
B
APPLICATIONS
Solve each triangle in Problems 21–32. If a problem has no
solution, say so.
37. Coast Guard. Two lookout posts, A and B (10.0 miles
apart), are established along a coast to watch for illegal
ships coming within the 3-mile limit. If post A reports a
ship S at angle BAS 37°30 and post B reports the same
ship at angle ABS 20°0, how far is the ship from post A?
How far is the ship from the shore (assuming the shore is
along the line joining the two observation posts)?
21. 118.3°, 12.2°, b 17.3 feet
22. 27.5°, 54.5°, a 9.27 inches
23. 67.7°, 54.2°, b 123 meters
24. 122.7°, 34.4°, b 18.3 kilometers
38. Fire Lookout. A fire at F is spotted from two fire lookout
stations, A and B, which are 10.0 miles apart. If station B reports the fire at angle ABF 53°0 and station A reports the
fire at angle BAF 28°30, how far is the fire from station
A? From station B?
25. 46.5°, a 7.9 millimeters, b 13.1 millimeters
26. 26.3°, a 14.7 inches, b 35.2 inches
27. 38.9°, a 30.0 inches, b 42.7 inches
28. 27.3°, a 135 centimeters, b 244 centimeters
★
29. 123.2°, a 101 yards, b 152 yards
30. 137.3°, a 13.9 meters, b 19.1 meters
31. 29°30, a 43.2 millimeters, b 56.5 millimeters
39. Natural Science. The tallest trees in the world grow in
Redwood National Park in California; they are taller than a
football field is long. Find the height of one of these trees,
given the information in the figure. (The 100-foot measurement is accurate to 3 significant digits.)
32. 33°50, a 673 meters, b 1,240 meters
C
33. Let 42.3° and b 25.2 centimeters. Determine a value
k so that if 0 a k, there is no solution; if a k, there is
one solution; and if k a b, there are two solutions.
34. Let 37.3° and b 42.8 centimeters. Determine a value
k so that if 0 a k, there is no solution; if a k, there is
one solution; and if k a b, there are two solutions.
3710
440
100 feet
35. Mollweide’s equation,
(a b) cos
c sin
2
2
is often used to check the final solution of a triangle, since
all six parts of a triangle are involved in the equation. If the
left side does not equal the right side after substitution, then
★
40. Surveying. To measure the height of Mt. Whitney in California, surveyors used a scheme like the one shown in the
figure in Problem 39. They set up a horizontal baseline
2,000 feet long at the foot of the mountain and found the
7-1
Law of Sines
515
angle nearest the mountain to be 43°5; the angle farthest
from the mountain was found to be 38°0. If the baseline
was 5,000 feet above sea level, how high is Mt. Whitney
above sea level?
41. Engineering. A 4.5-inch piston rod joins a piston to a 1.5inch crankshaft (see figure). How far is the base of the piston from the center of the crankshaft (distance d ) when the
rod makes an angle of 9° with the centerline? There are two
4.5 inches
Piston
1.5 inches
9
Crankshaft
d
★
46. Surveying. Find the height of the tree in Problem 45 if the
shadow length is 157 feet and, relative to the horizontal, the
hill slopes 11.0° and the angle of elevation of the sun is
42.0°.
★
47. Life Science. A cross section of the cornea of an eye, a circular arc, is shown in the figure. Find the arc radius R and
the arc length s, given the chord length C 11.8 millimeters and the central angle 98.9°.
42. Engineering. Repeat Problem 41 if the piston rod is 6.3
inches, the crankshaft is 1.7 inches, and the angle is 11°.
43. Astronomy. The orbits of the Earth and Venus are approximately circular, with the sun at the center. A sighting of
Venus is made from Earth, and the angle is found to be
18°40. If the radius of the orbit of the Earth is 1.495 108
kilometers and the radius of the orbit of Venus is 1.085 108 kilometers, what are the possible distances from the
Earth to Venus (see figure)?
R
44. Astronomy. In Problem 43, find the maximum angle .
[Hint: The angle is maximum when a straight line joining
the Earth and Venus is tangent to Venus’s orbit.]
C
R
Venus
s
Cornea
Sun
Venus
Earth
★
45. Surveying. A tree growing on a hillside casts a 102-foot
shadow straight down the hill (see figure). Find the vertical
height of the tree if, relative to the horizontal, the hill slopes
15.0° and the angle of elevation of the sun is 62.0°.
★
48. Life Science. Referring to the figure, find the arc radius R
and the arc length s, given the chord length C 10.2 millimeters and the central angle 63.2°.
★
49. Surveying. The procedure illustrated in Problems 39 and
40 is used to determine an inaccessible height h when a
baseline d on a line perpendicular to h can be established
(see the figure on the next page) and the angles and can
be measured. Show that
hd
sin sin sin ( )
516
h d sin csc ( ) tan h
h
d
★★
50. Surveying. The layout in the figure at right is used to determine an inaccessible height h when a baseline d in a
plane perpendicular to h can be established and the angles
, , and can be measured. Show that
SECTION
7-2
d
Law of Cosines
• Law of Cosines Derivation
• Solving the SAS Case
• Solving the SSS Case
If in a triangle two sides and the included angle are given (SAS) or three sides are
given (SSS), the law of sines cannot be used to solve the triangle—neither case
involves an angle and its opposite side (Fig. 1). Both cases can be solved starting
with the law of cosines, which is the subject matter for this section.
FIGURE 1
b
a
b
c
c
(a) SAS case
• Law of Cosines
(b) SSS case
Theorem 1 states the law of cosines.
Derivation
Theorem 1
Law of Cosines
a2 b2 c2 2bc cos b
a
c
b2 a2 c2 2ac cos c2 a2 b2 2ab cos All three equations
say essentially the
same thing.
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