Download SECTION 8-2 Law of Cosines

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8 Additional Topics in Trigonometry
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
8-2
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.
8-2 Law of Cosines
589
The law of cosines is used to solve triangles, given:
1. Two sides and the included angle (SAS)
2. Three sides (SSS)
We will establish the first equation in Theorem 1. The other two equations then
can be obtained from this one simply by relabeling the figure. We start by locating a
triangle in a rectangular coordinate system. Figure 2 shows three typical triangles.
For an arbitrary triangle located as in Figure 2, the distance-between-two-points
formula is used to obtain
a (hc)2 (k 0)2
a2 (h c)2 k2
Square both sides.
h2 2hc c2 k2
FIGURE 2 Three representative
triangles.
a
b
k
c (c, 0)
(h, k)
(h, k)
(h, k)
b
(1)
k
a
(a)
a
b
(c, 0)
h
c
h
k
(b)
(h, 0)
h0
c (c, 0)
(c)
From Figure 2, we note that
b2 h2 k2
Substituting b2 for h2 k2 in equation (1), we obtain
a2 b2 c2 2hc
(2)
But
cos h
b
h b cos Thus, by replacing h in equation (2) with b cos , we reach our objective:
a2 b2 c2 2bc cos [Note: If is acute, then cos is positive; if is obtuse, then cos is negative.]
• Solving the
SAS Case
For the SAS case, start by using the law of cosines to find the side opposite the given
angle. Then use either the law of cosines or the law of sines to find a second angle.
Because of the simpler computations, the law of sines will generally be used to find
the second angle.
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8 Additional Topics in Trigonometry
EXPLORE-DISCUSS 1
After using the law of cosines to find the side opposite the angle for a SAS case,
the law of sines is used to find a second angle. Figure 2 (a) shows that there are
two choices for a second angle.
(A) If the given angle is obtuse, can either of the remaining angles be obtuse?
Explain.
(B) If the given angle is acute, then one of the remaining angles may or may not
be obtuse. Explain why choosing the angle opposite the shorter side guarantees the selection of an acute angle.
(C) Starting with (sin )/a (sin )/b, show that
a sinb sin1
(1)
(D) Explain why equation (1) gives us the correct angle only if is acute.
The above discussion leads to the following strategy for solving the SAS case:
Strategy for Solving the SAS Case
EXAMPLE 1
Step
Find
Method
1.
Side opposite given angle
Law of cosines
2.
Second angle (Find the angle
opposite the shorter of the two
given sides—this angle will
always be acute.)
Law of sines
3.
Third angle
Subtract the sum of the measures
of the given angle and the angle
found in step 2 from 180°.
Solving the SAS Case
Solve the triangle in Figure 3.
FIGURE 3
10.3 cm
b
32.4
6.45 cm
8-2 Law of Cosines
591
Solution
Solve for b
Use the law of cosines:
b2 a2 c2 2ac cos Solve for b.
b a c 2ac cos 2
2
(10.3)2 (6.45)2 2(10.3)(6.45) cos 32.4°
5.96 cm
Solve for Since side c is shorter than side a, must be acute, and the law of sines is used to
solve for .
sin sin c
b
sin Solve for sin .
c sin b
sin1
sin1
Solve for .
b sin 32.4°
6.45 5.96
c sin Since is acute, the inverse sine
function gives us directly.
35.4°
Solve for 180° ( )
180° (32.4° 35.4°) 112.2°
Matched Problem 1
• Solving the
SSS Case
EXPLORE-DISCUSS 2
Solve the triangle with 77.5°, b 10.4 feet, and c 17.7 feet.
Starting with three sides of a triangle, the problem is to find the three angles. Subsequent calculations are simplified if we solve for the obtuse angle first, if present. The
law of cosines is used for this purpose. A second angle, which must be acute, can be
found using either law, although computations are usually simpler with the law of sines.
(A) Starting with a2 b2 c2 2bc cos , show that
cos1
a2 b2 c2
2bc
(2)
(B) Does equation (2) give us the correct angle irrespective of whether is
acute or obtuse? Explain.
The above discussion leads to the following strategy for solving the SSS case.
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8 Additional Topics in Trigonometry
Strategy for Solving the SSS Case
EXAMPLE 2
Step
Find
Method
1.
Angle opposite longest
side—this will take care of
an obtuse angle, if present.
Law of cosines
2.
Either of the remaining angles,
which will be acute (why?)
Law of sines
3.
Third angle
Subtract the sum of the measures
of the angles found in steps 1 and 2
from 180°.
Solving the SSS Case
Solve the triangle with a 27.3 meters, b 17.8 meters, and c 35.2 meters.
Solution
Three sides of the triangle are given and we are to find the three angles. This is the
SSS case.
Sketch the triangle (Fig. 4) and use the law of cosines to find the largest angle,
then use the law of sines to find one of the two remaining acute angles.
FIGURE 4
17.8 m
27.3 m
35.2 m
Since is the largest angle, we solve for it first using the law of cosines.
Solve for c2 a2 b2 2ab cos cos a2 b2 c2
2ab
cos1
cos1
Solve for cos .
a2 b2 c2
2ab
(27.3)2 (17.8)2 (35.2)2
2(27.3)(17.8)
Solve for .
100.5°
Solve for We now solve for either or , using the law of sines. We choose .
sin sin a
c
8-2 Law of Cosines
sin a sin 27.3 sin 100.5
c
35.2
sin1
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Solve for sin .
sin 100.5
27.3 35.2
Solve for .
49.7°
is acute.
180°
Solve for 180° ( )
180° (49.7° 100.5°)
29.8°
Matched Problem 2
EXAMPLE 3
Solve the triangle with a 1.25 yards, b 2.05 yards, and c 1.52 yards.
Finding the Side of a Regular Polygon
If a seven-sided regular polygon is inscribed in a circle of radius 22.8 centimeters,
find the length of one side of the polygon.
Solution
FIGURE 5
Sketch a figure (Fig. 5) and use the law of cosines:
Actually, you
only need to
sketch the
triangle:
22.8
360
7
22.8
d
d 2 22.82 22.82 2(22.8)(22.8) cos
d
2(22.8)2 2(22.8)2 cos
360°
7
360°
7
19.8 centimeters
Matched Problem 3
If an 11-sided regular polygon is inscribed in a circle with radius 4.63 inches, find
the length of one side of the polygon.
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8 Additional Topics in Trigonometry
Answers to Matched Problems
1. a 18.5 ft, 33.3°, 69.2°
2. 37.4°, 95.0°, 47.6°
EXERCISE
3. 2.61 in.
8-2
The labeling in the figure below is the convention we will
follow in this exercise set. Your answers to some problems
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.
12. a 31.5 meters, b 29.4 meters, c 33.7 meters
Problems 13–26 represent a variety of problems involving
both the law of sines and the law of cosines. Solve each
triangle. If a problem does not have a solution, say so.
13. 92.6°, 88.9°, a 15.2 centimeters
b
a
c
A
1. Referring to the figure above, if 47.3°, b 11.7 centimeters, and c 6.04 centimeters, which of the two angles, or , can you say for certain is acute and why?
2. Referring to the figure above, if 93.5°, b 5.34 inches,
and c 8.77 inches, which of the two angles, or , can
you say for certain is acute and why?
Solve each triangle in Problems 3–6.
14. 79.4°, 102.3°, a 6.4 millimeters
15. 126.2°, a 13.8 inches, c 12.5 inches
16. 19.1°, a 16.4 yards, b 28.2 yards
17. a 23.4 meters, b 6.9 meters, c 31.3 meters
18. a 86 inches, b 32 inches, c 53 inches
19. 38.4°, a 11.5 inches, b 14.0 inches
20. 66.4°, b 25.5 meters, c 25.5 meters
21. a 32.9 meters, b 42.4 meters, c 20.4 meters
22. a 10.5 centimeters, b 5.23 centimeters, c 8.66 centimeters
23. 58.4°, b 7.23 meters, c 6.54 meters
24. 46.7°, a 18.1 meters, b 22.6 meters
3. 71.2°, b 5.32 yards, c 5.03 yards
25. 39.8°, a 12.5 inches, b 7.31 inches
4. 57.3°, a 6.08 centimeters, c 5.25 centimeters
26. 47.9°, b 35.2 inches, c 25.5 inches
5. 120°20, a 5.73 millimeters, b 10.2 millimeters
6. 135°50, b 8.44 inches, c 20.3 inches
B
7. Referring to the figure at the beginning of the exercise, if
a 13.5 feet, b 20.8 feet, and c 8.09 feet, then if the
triangle has an obtuse angle, which angle must it be and
why?
8. Suppose you are told that a triangle has sides a 12.5 centimeters, b 25.3 centimeters, and c 10.7 centimeters.
Explain why the triangle has no solution.
Solve each triangle in Problems 9–12 if the triangle has a
solution. Use decimal degrees for angle measure.
9. a 4.00 meters, b 10.2 meters, c 9.05 meters
10. a 10.5 miles, b 20.7 miles, c 12.2 miles
11. a 6.00 kilometers, b 5.30 kilometers, c 5.52
kilometers
C
27. Show, using the law of cosines, that if 90°, then c2 a2 b2 (the Pythagorean theorem).
28. Show, using the law of cosines, that if c2 a2 b2, then
90°.
29. Show that for any triangle,
a2 b2 c2 cos cos cos 2abc
a
b
c
30. Show that for any triangle,
a b cos c cos 31. Give a solution to Example 3 that does not use the law of
cosines by showing that
d
360˚
22.8 sin
.
2
14
8-2 Law of Cosines
32. Show that the length d of one side of an n-sided regular
polygon, inscribed in a circle of radius r, is given by
d 2r sin
★
180˚
.
n
41. Analytic Geometry. If point A in the figure has coordinates
(3, 4) and point B has coordinates (4, 3), find the radian
measure of angle to three decimal places.
y
APPLICATIONS
A
B
33. Surveying. To find the length AB of a small lake, a surveyor measured angle ACB to be 96°, AC to be 91 yards,
and BC to be 71 yards. What is the approximate length of
the lake?
x
0
★
42. Analytic Geometry. If point A in the figure has coordinates
(4, 3) and point B has coordinates (5, 1), find the radian
measure of angle to three decimal places.
★
43. Engineering. Three circles of radius 2.03, 5.00, and 8.20
centimeters are tangent to one another (see figure). Find the
three angles formed by the lines joining their centers (to the
nearest 10).
C
A
595
B
35. Geometry. Two adjacent sides of a parallelogram meet at
an angle of 35°10 and have lengths of 3 and 8 feet. What is
the length of the shorter diagonal of the parallelogram (to 3
significant digits)?
★
36. Geometry. What is the length of the longer diagonal of the
parallelogram in Problem 35 (to 3 significant digits)?
C
2.8 cm
38. Search and Rescue. At noon, two search planes set out
from San Francisco to find a downed plane in the ocean.
Plane A travels due west at 400 miles per hour, and plane B
flies northwest at 500 miles per hour. At 2 P.M. plane A spots
the survivors of the downed plane and radios plane B to
come and assist in the rescue. How far is plane B from plane
A at this time (to 3 significant digits)?
40. Geometry. Find the perimeter of a nine-sided regular polygon inscribed in a circle of radius 7.09 centimeters.
44. Engineering. Three circles of radius 2.00, 5.00, and 8.00
inches are tangent to each other (see figure). Find the three
angles formed by the lines joining their centers (to the nearest 10).
45. Geometry. A rectangular solid has sides as indicated in the
figure. Find CAB to the nearest degree.
37. Navigation. Los Angeles and Las Vegas are approximately
200 miles apart. A pilot 80 miles from Los Angeles finds
that she is 6°20 off course relative to her start in Los Angeles. How far is she from Las Vegas at this time? (Compute the answer to 3 significant digits.)
39. Geometry. Find the perimeter of a pentagon inscribed in a
circle of radius 12.6 meters.
34. Surveying. Suppose the figure for this problem represents
the base of a large rock outcropping on a farmer’s land. If a
surveyor finds ACB 110°, AC 85 meters, and
BC 73 meters, what is the approximate length (to one
decimal place) of the outcropping?
B
A
4.3 cm
8.1 cm
46. Geometry. Referring to the figure, find ACB to the nearest degree.
★
47. Space Science. For communications between a space shuttle and the White Sands tracking station in southern New
Mexico, two satellites are placed in geostationary orbit,
130° apart relative to the center of the Earth and 22,300
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8 Additional Topics in Trigonometry
miles above the surface of the Earth (see figure). (A satellite in geostationary orbit remains stationary above a fixed
point on the surface of the Earth.) Radio signals are sent
from the tracking station by way of the satellites to the shuttle, and vice versa. This system allows the tracking station
to be in contact with the shuttle over most of the Earth’s
surface. How far to the nearest 100 miles is one of the geostationary satellites from the White Sands tracking station
W? The radius of the Earth is 3,964 miles.
S
★
48. Space Science. A satellite S, in circular orbit around the
Earth, is sighted by a tracking station T (see figure). The
distance TS is determined by radar to be 1,034 miles, and
the angle of elevation above the horizon is 32.4°. How high
is the satellite above the Earth at the time of the sighting?
The radius of the Earth is 3,964 miles.
S
T
Horizon
S
R
W
Earth
C
C
SECTION
8-3
Geometric Vectors
•
•
•
•
Geometric Vectors and Vector Addition
Velocity Vectors
Force Vectors
Resolution of Vectors into Vector Components
Many physical quantities, such as length, area, or volume, can be completely specified by a single real number. Other quantities, such as directed distances, velocities,
and forces, require for their complete specification both a magnitude and a direction. The former are often called scalar quantities, and the latter are called vector
quantities.
In this section we limit our discussion to the intuitive idea of geometric vectors
in a plane. In Section 8-4 we introduce algebraic vectors, a first step in the generalization of a concept that has far-reaching consequences. Vectors are widely used in
many areas of science and engineering.
• Geometric Vectors
and Vector Addition
P
v
O
, or v.
FIGURE 1 Vector OP
A line segment to which a direction has been assigned is called a directed line segment. A geometric vector is a directed line segment and is represented by an arrow
(see Fig. 1). A vector with an initial point O and a terminal point P (the end with
. Vectors are also denoted by a boldface letter, such
the arrowhead) is denoted by OP
as v. Since it is difficult to write boldface on paper, we suggest that you use an arrow
over a single letter, such as v , when you want the letter to denote a vector.
, denoted by OP
, v
or v, is the length of the
The magnitude of the vector OP
directed line segment. Two vectors have the same direction if they are parallel and
point in the same direction. Two vectors have opposite direction if they are parallel
and point in opposite directions. The zero vector, denoted by 0 or 0, has a magni-