5 Oblique Triangles, the Law of Sines
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9.5 Oblique Triangles, the Law of Sines
LEARNING TIP
Note that there are actually three
equations combined in Eq. (9.8). Of
these, we use the one with three
known parts of the triangle, and we
find the fourth part. In finding the
complete solution of a triangle, it
may be necessary to use two of the
three equations.
283
Another form of the law of sines can be obtained by equating the reciprocals of each of
the fractions in Eq. (9.8). The law of sines is a statement of proportionality between the
sides of a triangle and the sines of the angles opposite them.
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Now, we see how the law of sines is used in the solution of a triangle in which two
angles and one side are known. If two angles are known, the third may be found from
the fact that the sum of the angles in a triangle is 180°. At this point, we must be able to
find the ratio between the given side and the sine of the angle opposite it. Then, by use
of the law of sines, we may find the other two sides.
E X A M P L E 1 Case 1: Two angles and one side
Given c = 6.00, A = 60.0°, and B = 40.0°, find a, b, and C.
First, we can see that
C = 180.0° - 160.0° + 40.0°2 = 80.0°
We now know side c and angle C, which allows us to use Eq. (9.8). Therefore, using
the equation relating a, A, c, and C, we have
B
c
6.00 40.0° a
60.0°
A
C
b
a
6.00
=
sin 60.0°
sin 80.0°
or a =
A
C
6.00 sin 60.0°
= 5.28
sin 80.0°
Now, using the equation relating b, B, c, and C, we have
c
Fig. 9.47
Practice Exercise
1. In Example 1, change the value of B to
65.0°, and then find a.
LEARNING TIP
The solution of a triangle can be
checked approximately by noting
that the smallest angle is opposite
the shortest side, and the largest
angle is opposite the longest side.
Note that this is the case in Example
1, where b (shortest side) is opposite
B (smallest angle), and c (longest
side) is opposite C (largest angle).
b
6.00
=
sin 40.0°
sin 80.0°
or b =
B
C
6.00 sin 40.0°
= 3.92
sin 80.0°
Thus, a = 5.28, b = 3.92, and C = 80.0°. See Fig. 9.47. We could also have used the
form of Eq. (9.8) relating a, A, b, and B in order to find b, but any error in calculating a
would make b in error as well. Of course, any error in calculating C would make both a
and b in error.
■
E X A M P L E 2 Case 1: Two angles and one side
Solve the triangle with the following given parts: a = 63.71, A = 56.29°, and
B = 97.06°. See Fig. 9.48.
From the figure, we see that we are to find angle C and sides b and c. We first determine angle C:
C = 180° - 1A + B2 = 180° - 156.29° + 97.06°2
= 26.65°
Noting the three angles, we know that c is the shortest side (C is the smallest angle)
and b is the longest side (B is the largest angle). This means that the length of a is
between c and b, or c 6 63.71 and b 7 63.71. Now using the ratio a>sin A of Eq. (9.8)
(the law of sines) to find sides b and c, we have
a
b
63.71
=
sin 97.06°
sin 56.29°
B
c 97.06°
56.29°
A
b
B
63.71
Fig. 9.48
C
63.71 sin 97.06°
= 76.01
sin 56.29°
or c =
63.71 sin 26.65°
= 34.35
sin 56.29°
A
a
c
63.71
=
sin 26.65°
sin 56.29°
C
or b =
A
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CHAPTER 9
Vectors and Oblique Triangles
Thus, b = 76.01, c = 34.35, and C = 26.65°. Note that c 6 a and b 7 a, as
expected.
■
If the given information is appropriate, the law of sines may be used to solve applied
problems. The following example illustrates the use of the law of sines in such a
problem.
E X A M P L E 3 Case 1: Application
Two observers A and B sight a helicopter due east. The observers are 1540 m apart, and
the angles of elevation they each measure to the helicopter are 32.0° and 44.0°, respectively. How far is observer A from the helicopter? See Fig. 9.49.
Letting H represent the position of the helicopter, we see that angle B within the
triangle ABH is 180° - 44.0° = 136.0°. This means that the angle at H within the
triangle is
■ The first successful helicopter was made in
the United States by Igor Sikorsky in 1939.
H = 180° - 132.0° + 136.0°2 = 12.0°
Now, using the law of sines to find required side b, we have
required side
opposite
known angle
H
b
A
44.0°
32.0°
b
1540
=
sin 136.0°
sin 12.0°
known side
opposite
known angle
or
1540 m B
b =
Fig. 9.49
1540 sin 136.0°
= 5150 m
sin 12.0°
Thus, observer A is about 5150 m from the helicopter.
■
CASE 2: TWO SIDES AND THE ANGLE OPPOSITE ONE OF THEM
For a triangle in which we know two sides and the angle opposite one of the given
sides, the solution will be either one triangle, or two triangles, or even possibly no triangle. The following examples illustrate how each of these results is possible.
E X A M P L E 4 Case 2: Two sides and angle opposite
Solve the triangle with the following given parts: a = 40.0, b = 60.0, and A = 30.0°.
First, make a good scale drawing (Fig. 9.50(a)) by drawing angle A and measuring
off 60 for b. This will more clearly show that side a = 40.0 will intersect side c at
either position B or B′. This means there are two triangles that satisfy the given values.
Using the law of sines, we solve the case for which B is an acute angle:
b = 60.0
30.0°
A
C
a = 40.0
a = 40.0
B'
B
30.0°
A
b = 60.0
30.0°
A
c'
C'
a = 40.0
B'
(c)
Fig. 9.50
B
c
(b)
Side a reaches
B at either
of two points
(a)
a = 40.0
b = 60.0
60.0
40.0
60.0 sin 30.0°
=
or sin B =
sin B
sin 30.0°
40.0
60.0 sin 30.0°
B = sin - 1 a
b = 48.6°
40.0
C = 180° - 130.0° + 48.6° 2 = 101.4°
Therefore, B = 48.6° and C = 101.4°. Using the law of sines again to find
c, we have
c
40.0
=
sin 101.4°
sin 30.0°
40.0 sin 101.4°
c =
sin 30.0°
= 78.4
Thus, B = 48.6°, C = 101.4°, and c = 78.4. See Fig. 9.50(b).
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285
9.5 Oblique Triangles, the Law of Sines
The other solution is the case in which B′, opposite side b, is an obtuse angle.
Therefore,
B′ =
=
C′ =
=
180° - B = 180° - 48.6°
131.4°
180° - 130.0° + 131.4° 2
18.6°
Using the law of sines to find c′, we have
c′
40.0
=
sin 18.6°
sin 30.0°
40.0 sin 18.6°
c′ =
sin 30.0°
= 25.5
This means that the second solution is B′ = 131.4°, C′ = 18.6°, and c′ = 25.5. See
Fig. 9.50(c).
■
E X A M P L E 5 Case 2: Possible solutions
In Example 4, if a were 60.0, only one solution would result. In this case, side a would intercept side c at B. It also intercepts the extension of side c, but this would require that angle A
not be included in the triangle (see Fig. 9.51). Thus, only one solution may result if a 7 b.
In Example 4, there would be no solution if side a were not at least 30.0. If this were
the case, side a would not be long enough to even touch side c. It can be seen that a
must at least equal b sin A. If it is just equal to b sin A, there is one solution, a right triangle. See Fig. 9.52.
30.0°
A'
B
Practice Exercise
2. Determine which of the four possible
solution types occurs if a = 28, b = 48,
and A = 30°.
60.0
A
Side b reaches A
but too long for
a second A
30.0°
A
Fig. 9.52
A
Summary of Solutions:
Two Sides and the Angle Opposite One of Them
1. No solution if a 6 b sin A. See Fig. 9.53(a).
2. A right triangle solution if a = b sin A. See Fig. 9.53(b).
3. Two solutions if b sin A 6 a 6 b. See Fig. 9.53(c).
4. One solution if a 7 b. See Fig. 9.53(d).
a
b
A
(a)
Just
touches
■
Summarizing the results for Case 2 as illustrated in Examples 4 and 5, we make the
following conclusions. Given sides a and b and angle A (assuming here that a and A
1A 6 90°2 are corresponding parts), we have the following summary of solutions for
Case 2.
C'
C
b
30.0
B
Fig. 9.51
Ambiguous Case
C
b > 60.0
a = 60.0
b
a
A
b
a
B
(b)
(c)
Fig. 9.53
A
B'
a
b
a
A
(d)
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CHAPTER 9
Vectors and Oblique Triangles
LEARNING TIP
Note that in order to have two solutions, we must know two sides and
the angle opposite one of the sides,
and the shorter side must be opposite
the known angle.
If there is no solution, the calculator will indicate an error. If the solution is a right triangle, the calculator
will show an angle of exactly 90° (no
extra decimal digits will be displayed).
■ See the chapter introduction
v
pa
=
30
0
He km
/h
ad
in
g
North
θ
For the reason that two solutions may result from it, Case 2 is called the ambiguous
case. The following example illustrates Case 2 in an applied problem.
E X A M P L E 6 Case 2: Application
Edmonton is 35.2° north of east of Vancouver. What should be the heading of a plane
from Vancouver to Edmonton if the wind is from the west at 40.0 km>h and the plane’s
speed with respect to the air is 300 km>h?
The heading should be set so that the resultant of the plane’s velocity with respect to
the air vpa and the velocity of the wind vw will be in the direction from Vancouver to
Edmonton. This means that the resultant velocity vpg of the plane with respect to the
ground must be at an angle of 35.2° north of east of Vancouver.
Using the given information, we draw the vector triangle shown in Fig. 9.54. In the
triangle, we know that the angle at Edmonton is 35.2° by noting the alternate-interior
angles (see page 58). By finding u, the required heading can be found. There can be
only one solution, since vpa 7 vw. Using the law of sines, we have
v w = 40.0 km/h
Edmonton
known side
opposite
required angle
vpg
35.2°
sin u =
East
Vancouver
40.0
300
=
sin u
sin 35.2°
Fig. 9.54
known side
opposite
known angle
40.0 sin 35.2°
,
300
u = 4.4°
Therefore, the heading should be 35.2° + 4.4° = 39.6° north of east. Compare this
example with Example 3 on page 278.
■
If we try to use the law of sines for Case 3 or Case 4, we find that we do not have
enough information to complete any of the ratios. These cases can, however, be solved
by the law of cosines as shown in the next section.
E X A M P L E 7 Cases 3 and 4 not solvable by the law of sines
Given (Case 3) two sides and the included the angle of a triangle a = 2, b = 3, C = 45°,
and (Case 4) the three sides a = 5, b = 6, c = 7, we set up the ratios
(Case 3)
2
3
c
5
6
7
=
=
, and (Case 4)
=
=
sin A
sin B
sin 45°
sin A
sin B
sin C
The solution cannot be found since each of the three possible equations in either Case 3
or Case 4 contains two unknowns.
■
E XE R C I S E S 9 .5
In Exercises 1 and 2, solve the resulting triangles if the given changes
are made in the indicated examples of this section.
7. a = 4.601, b = 3.107, A = 18.23°
1. In Example 2, solve the triangle if the value of B is changed to
82.94°.
9. b = 7751, c = 3642, B = 20.73°
2. In Example 4, solve the triangle if the value of b is changed to
70.0.
In Exercises 3–22, solve the triangles with the given parts.
8. b = 362.2, c = 294.6, B = 69.37°
10. a = 150.4, c = 250.9, C = 76.43°
11. b = 0.0742, B = 51.0°, C = 3.40°
12. c = 729, B = 121.0°, C = 44.2°
13. a = 63.8, B = 58.4°, C = 22.2°
3. a = 45.7, A = 65.0°, B = 49.0°
14. a = 0.130, A = 55.2°, B = 117.5°
4. b = 3.07, A = 26.0°, C = 120.0°
15. b = 4384, B = 47.43°, C = 64.56°
5. c = 4380, A = 37.4°, B = 34.6°
16. b = 283.2, B = 13.79°, C = 76.38°
6. a = 93.2, B = 0.9°, C = 82.6°
17. a = 5.240, b = 4.446, B = 48.13°
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9.5 Oblique Triangles, the Law of Sines
18. a = 89.45, c = 37.36, C = 15.62°
19. b = 2880, c = 3650, B = 31.4°
56 km
Golden, BC
20. a = 0.841, b = 0.965, A = 57.1°
287
Lake Louise, AB
66°
21. a = 450, b = 1260, A = 64.8°
90 km
22. a = 20, c = 10, C = 30°
In Exercises 23–41, use the law of sines to solve the given problems.
23. Two angles of a triangle measure 29.0° and 57.0°. The longest
side is 52.0 cm long. What is the length of the shortest side?
25. The loading ramp at a package delivery service is 3.2 m long and
makes a 22.5° angle with the level ground. If this ramp is replaced
with one that is 7.5 m long, what angle does the new ramp make
with the ground?
Radium, BC
Fig. 9.57
31. Find the length of Cordelia St. in Brisbane, Queensland, Australia,
from Fig. 9.58.
T1
Boundary St.
t.
30.0°
42.0°
aS
eli
28. Two ropes hold a 175-N crate as shown in Fig. 9.55. Find the tensions T1 and T2 in the ropes. (Hint: Move the vectors so that they
are tail to head to form a triangle. The vector sum T1 + T2 must
equal 175 N for equilibrium. See page 278.)
rd
27. The floor of the Bastion in Nanaimo, British Columbia, is a regular octagon, 2.3 m on each side. Find the greatest distance across
the floor (that is, find the length of the longest diagonal).
Co
26. The longest side of a triangular parcel of land is 172 m long, and
the shortest side is 105 m long. If the largest angle is 82.0°, what
is the length of the third side?
920 m
24. Two angles of a triangle measure 45.0° and 55.0°. The side opposite the 55.0° angle is 8.75 cm long. What is the length of the
longest side?
810 m
Fig. 9.58
49
Vulture St.
32. When an airplane is landing on a an 2510-m runway, the angles
of depression to the ends of the runway are 10.0° and 13.5°. How
far is the plane from the near end of the runway?
33. Find the total length of the path of the laser beam that is shown in
Fig. 9.59.
T2
6.25 cm
Crate
108.3°
Fig. 9.55
31.8°
175 N
Fig. 9.59
29. Find the tension T in the left guy wire attached to the top of the
tower shown in Fig. 9.56. (Hint: The horizontal components of
the tensions must be equal and opposite for equilibrium. See page
278. Thus, move the tension vectors tail to head to form a triangle
with a vertical resultant. This resultant equals the upward force at
the top of the tower for equilibrium. This last force is not shown
and does not have to be calculated.)
T
105.6°
Fig. 9.56
Reflectors
850 N
35.7°
34. In widening a highway, it is necessary for a construction crew to
cut into the bank along the highway. The present angle of elevation of the straight slope of the bank is 23.0°, and the new angle is
to be 38.5°, leaving the top of the slope at its present position. If
the slope of the present bank is 66.0 m long, how far horizontally
into the bank at its base must they dig?
35. A communications satellite is directly above the extension of a
line between receiving towers A and B. It is determined from
radio signals that the angle of elevation of the satellite from tower
A is 89.2°, and the angle of elevation from tower B is 86.5°. See
Fig. 9.60. If A and B are 1290 km apart, how far is the satellite
from A? (Ignore the curvature of the earth.)
Satellite
30. Find the distance from Golden to Radium, from Fig. 9.57. The
route that joins these landmarks across three national parks is
known as the Golden Triangle.
Fig. 9.60
86.5°
89.2°
B 1290 km A
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CHAPTER 9
Vectors and Oblique Triangles
36. An astronaut on the moon drives a lunar rover 16 km in the direction 60.0° north of east from the base. (a) Through what angle
must the rover then be turned so that by driving 12 km farther the
astronaut can turn again to return to base along a north-south line?
(b) How long is the last leg of the trip? (c) Can the astronaut make
it back to base if the maximum range of the rover is 40 km?
37. A boat owner wishes to cross a river 2.60 km wide and go directly
to a point on the opposite side 1.75 km downstream. The boat
goes 8.00 km>h in still water, and the stream flows at 3.50 km>h.
What should the boat’s heading be?
40. Point P on the mechanism shown in Fig. 9.61 is driven back and
forth horizontally. If the minimum value of angle u is 32.0°, what
is the distance between extreme positions of P? What is the maximum possible value of angle u?
P
u
Fig. 9.61
41. The floor of the Winnipeg Art Gallery is in the shape of a triangle. Find the length of the side facing Memorial Boulevard, from
Fig. 9.62.
Colony St.
114 m
0 m ve.
61.
ry A
Ma
39. A hillside is inclined at 23° with the horizontal. From a given point
on the slope, it has been found that a vein of gold is 55 m directly
below. At what point downhill and at what angle below the hillside
slope must a straight 65-m shaft be dug to reach the vein?
24.5 cm
St.
38. A motorist travelling along a level highway at 75 km>h directly
toward a mountain notes that the angle of elevation of the mountain top changes from about 20° to about 30° in a 20-min period.
How much closer on a direct line did the mountain top become?
36.0 cm
82.6°
.
lvd
lB
a
ori
m
Me
Fig. 9.62
Answers to Practice Exercises
1. a = 6.34
2. Two solutions
The Law of Cosines
9.6
-BXPG$PTJOFT t $BTF5XP4JEFTBOE
*ODMVEFE"OHMF t $BTF5ISFF4JEFT t
Summary of Solving Oblique Triangles
C
b
A
a
h
x
B
c
As noted in the last section, the law of sines cannot be used for Case 3 (two sides and
the included angle) and Case 4 (three sides). In this section, we develop the law of
cosines, which can be used for Cases 3 and 4. After finding another part of the triangle
using the law of cosines, we will often find it easier to complete the solution using the
law of sines.
Consider any oblique triangle—for example, either triangle shown in Fig. 9.63. For
each triangle, h>b = sin A, or h = b sin A. Also, using the Pythagorean theorem, we
obtain a2 = h2 + x2 for each triangle. Therefore (with (sin A)2 = sin2A),
a2 = b2 sin2 A + x2
(a)
C
b
a
A
c
In Fig. 9.63(a), note that 1c - x2 >b = cos A, or c - x = b cos A. Solving for x, we
have x = c - b cos A. In Fig. 9.63(b), c + x = b cos A, and solving for x, we have
x = b cos A - c. Substituting these relations into Eq. (9.9), we obtain
h
x
B
(9.9)
and
a2 = b2 sin2 A + (c - b cos A)2
a2 = b2 sin2 A + (b cos A - c)2
(9.10)
respectively. When expanded, these both give
(b)
a2 = b2 sin2 A + b2 cos2 A + c2 - 2bc cos A
= b2 1sin2 A + cos2 A2 + c2 - 2bc cos A
Fig. 9.63
(9.11)
Recalling the definitions of the trigonometric functions, we know that sin u = y>r
and cos u = x>r. Thus, sin2 u + cos2 u = 1y2 + x2 2 >r 2. However, x2 + y2 = r 2,
which means
sin2 u + cos2 u = 1
(9.12)
This equation is valid for any angle u, since we have made no assumptions as to the properties of u. Thus, by substituting Eq. (9.12) into Eq. (9.11), we arrive at the law of cosines:
Law of Cosines
a2 = b2 + c2 - 2bc cos A
(9.13)
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9.6 The Law of Cosines
289
Using the method above, we may also show that
b2 = a2 + c2 - 2ac cos B
c2 = a2 + b2 - 2ab cos C
$"4&5804*%&4"/%5)&*/$-6%&%"/(-&
If two sides and the included angle of a triangle are known, the forms of the law of
cosines show that we may directly solve for the side opposite the given angle. Then, as
noted earlier, the solution may be completed using the law of sines.
E X A M P L E 1 Case 3: Two sides and included angle
Solve the triangle with a = 45.0, b = 67.0, and C = 35.0°. See Fig. 9.64.
Since angle C is known, first solve for side c, using the law of cosines in the form
c2 = a2 + b2 - 2ab cos C. Substituting, we have
C
35.0°
67.0
45.0
unknown side opposite known angle
2
A
c = 45.02 + 67.02 - 2145.02 167.02 cos 35.0°
B
c
known sides
Fig. 9.64
c = 245.02 + 67.02 - 2145.02 167.02 cos 35.0° = 39.7
From the law of sines, we now have
Practice Exercise
1. In Example 1, change the value of a to
95.0, and find A and B.
COMMON ERROR
sides
45.0
67.0
39.7
opposite
=
=
sin A
sin B
sin 35.0°
angles
45.0 sin 35.0°
sin A =
,
A = 40.6°
39.7
Finally, rather than use the law of sines again, we obtain angle B by subtraction. We
find that B = 180° - (35.0° + 40.6°) = 104.4°. Therefore, c = 39.7, A = 40.6°,
and B = 104.4°.
■
If after finding side c we had solved for angle B rather than angle A, the calculator
would have shown 75.6°. This would have given us the value of the reference angle for
angle B, and not the correct value for the obtuse angle B (B = 180° – 75.6° = 104.4°).
Since only the largest angle of a triangle can be greater than 90°, we can avoid this
possible source of error if we first solve for the smaller unknown angle (the angle opposite
the shorter known side). The larger unknown angle can then be found by subtraction.
E X A M P L E 2 Case 3: Application
Two forces are acting on a bolt. One is a 78.0-N force acting horizontally to the right,
and the other is a force of 45.0 N acting upward to the right, 15.0° from the vertical.
Find the resultant force F. See Fig. 9.65.
Moving the 45.0-N vector to the right and using the lower triangle with the 105.0°
angle, the magnitude of F is
45.0 N
F
15.0°
45.0 N
105.0°
u
Bolt
78.0 N
Fig. 9.65
F = 278.02 + 45.02 - 2178.02 145.02 cos 105.0°
= 99.6 N
To find u, use the law of sines:
45.0
99.6
=
,
sin u
sin 105.0°
sin u =
45.0 sin 105.0°
99.6
This gives us u = 25.9°.
We can also solve this problem using vector components.
■
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CHAPTER 9
Vectors and Oblique Triangles
LEARNING TIP
The best procedure for Case 4 is to
find the largest angle first. This avoids
the ambiguous case if we switch to the
law of sines and there is an obtuse
angle. The largest angle is opposite the
longest side. Another procedure is to use
the law of cosines to find two angles.
CASE 4: THREE SIDES
Given the three sides of a triangle, we may solve for the angle opposite any side using
the law of cosines.
E X A M P L E 3 Case 4: Three sides
Solve the triangle with sides a = 49.33, b = 21.61, and c = 42.57. See Fig. 9.66.
Since the longest side is a = 49.33, first solve for angle A:
a2 = b2 + c2 - 2bc cos A
b2 + c2 - a2
21.612 + 42.572 - 49.332
cos A =
=
2bc
2(21.61)(42.57)
A = 94.81°
C
49.33
21.61
From the law of sines, we now have
A
B
42.57
49.33
21.61
42.57
=
=
sin A
sin B
sin C
Fig. 9.66
Therefore,
B = sin-1
21.61 sin 94.81°
= 25.88°
49.33
and C = 180° - 194.81° + 25.88°2 = 59.31°.
■
E X A M P L E 4 Case 4: Application
29.0 m
34.0 m
u
22.0 m
Fig. 9.67
■ Summary of solving an oblique triangle
using the law of sines or the law of cosines
Practice Exercise
2. Using Fig. 9.67, and only the data given
in Example 4, find the angle between the
guy wire and the hillside.
A vertical radio antenna is to be built on a hillside with a constant slope. A guy wire is
to be attached at a point 29.0 m up the antenna, and at a point 22.0 m from the base of
the antenna up the hillside. If the guy wire is 34.0 m long, what angle does the antenna
make with the hillside?
From Fig. 9.67, we can set up the equation necessary for the solution.
34.02 = 22.02 + 29.02 - 2(22.0)(29.0)cos u
u = cos - 1
22.02 + 29.02 - 34.02
= 82.4°
2(22.0)(29.0)
40-7*/(0#-*26&53*"/(-&4
Case 1: Two Angles and One Side
Find the unknown angle by subtracting the sum of the known angles
from 180°. Use the law of sines to find the unknown sides.
Case 2: Two Sides and the Angle Opposite One of Them
Use the known side and the known angle opposite it to find the angle
opposite the other known side. Find the third angle from the fact that
the sum of the angles is 180°. Use the law of sines to find the third side.
CAUTION: There may be two solutions. See page 285 for a summary
of Case 2 and the ambiguous case.
Case 3: Two Sides and the Included Angle
Find the third side by using the law of cosines. Find the smaller
unknown angle (opposite the shorter side) by using the law of sines.
Complete the solution using the fact that the sum of the angles is 180°.
Case 4: Three Sides
Find the largest angle (opposite the longest side) by using the law of
cosines. Find a second angle by using the law of sines. Complete the
solution by using the fact that the sum of the angles is 180°.
■
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9.6 The Law of Cosines
291
Other variations in finding the solutions can be used. For example, after finding the
third side in Case 3 or finding the largest angle in Case 4, the solution can be completed
by using the law of sines. All angles in Case 4 can be found by using the law of cosines.
The methods shown above are those normally used.
E XE RC IS ES 9 .6
In Exercises 1 and 2, solve the resulting triangles if the given changes
are made in the indicated examples of this section.
1. In Example 1, solve the triangle if the value of C is changed to
145°.
2. In Example 3, solve the triangle if the value of a is changed to
29.33.
In Exercises 3–22, solve the triangles with the given parts.
3. a = 6.00, b = 7.56, C = 54.0°
4. b = 87.3, c = 34.0, A = 130.0°
5. a = 4530, b = 924, C = 98.0°
25. On page 285, we saw that it was possible to be given two sides
and the angle opposite one of them, and find that there is no triangle that can have those parts. Is it possible in Case 3 or in Case 4
that there is no solution for a set of given parts?
26. Three circles of radii 24 cm, 32 cm, and 42 cm are externally tangent to each other (each is tangent to the other two). Find the largest angle of the triangle formed by joining their centres.
27. A nuclear submarine leaves its base and travels at 23.5 km>h. For
2.00 h, it travels along a course of 32.1° north of west. It then
turns an additional 21.5° north of west and travels for another
1.00 h. How far from its base is it?
28. The robot arm shown in Fig. 9.69 places packages on a conveyor
belt. What is the distance x?
6. a = 0.0845, c = 0.116, B = 85.0°
7. a = 39.53, b = 45.22, c = 67.15
8. a = 2.331, b = 2.726, c = 2.917
1.75 m
2.50 m
9. a = 385.4, b = 467.7, c = 800.9
102.0°
10. a = 0.2433, b = 0.2635, c = 0.1538
11. a = 320, b = 847, C = 158.0°
x
Fig. 9.69
12. b = 18.3, c = 27.1, A = 8.7°
Belt
29. Find the angle between the front legs and the back legs of the
folding chair shown in Fig. 9.70.
13. a = 2140, c = 428, B = 86.3°
14. a = 1.13, b = 0.510, C = 77.6°
15. b = 103.7, c = 159.1, C = 104.67°
16. a = 49.32, b = 54.55, B = 114.36°
17. a = 0.4937, b = 0.5956, c = 0.6398
cm
19. a = 723, b = 598, c = 158
64
53 c
m
18. a = 69.72, b = 49.30, c = 22.29
20. a = 1.78, b = 6.04, c = 4.80
21. a = 1500, A = 15°, B = 140°
24. Set up equations (do not solve) to solve the triangle in Fig. 9.68
by the law of cosines. Why is the law of sines easier to use?
30. In a baseball field, the four bases are at the vertices of a square
90.0 ft on a side. The pitching rubber is 60.5 ft from home plate.
See Fig. 9.71. How far is it from the pitching rubber to first base?
11 ft = 0.3048 m2
C
?
60.5 ft
a
10
20°
Fig. 9.68
A
30
Fig. 9.71
ft
23. For a triangle with sides a, b, and c opposite angles A, B, and C,
1b + c + a2 1b + c - a2
respectively, show that 1 + cos A =
.
2bc
48 cm
.0
In Exercises 23–40, use the law of cosines to solve the given problems.
Fig. 9.70
90
22. a = 17, b = 24, c = 42. Explain your answer.
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292
CHAPTER 9
Vectors and Oblique Triangles
31. A plane leaves an airport and travels 624 km due east. It then
turns toward the north and travels another 326 km. It then turns
again less than 180° and travels another 846 km directly back to
the airport. Through what angles did it turn?
32. The apparent depth of an object submerged in water is less than
its actual depth. A coin is actually 5.00 cm from an observer’s eye
just above the surface, but it appears to be only 4.25 cm. The real
light ray from the coin makes an angle with the surface that is 8.1°
greater than the angle the apparent ray makes. How much deeper
is the coin than it appears to be? See Fig. 9.72.
respect to the land in the direction of its destination. If the ferryboat’s heading is 23.6° from the direction of its destination, what
is the velocity of the current?
36. Find the distance across Georgian Bay from Nottawasaga Island
to Christian Island, from Fig. 9.74.
Cape Crocker
69 km
22.5°
72 km
4.25 cm
8.1°
Apparent
coin
5.00 cm
Fig. 9.74
Coin
Fig. 9.72
33. A nut is in the shape of a regular hexagon (six sides). If each side
is 9.53 mm, what opening on a wrench is necessary to tighten the
nut? See Fig. 9.73.
Christian Island
Nottawasaga Island
37. An air traffic controller sights two planes that are due east from
the control tower and headed toward each other. One is 15.8 km
from the tower at an angle of elevation of 26.4°, and the other is
32.7 km from the tower at an angle of elevation of 12.4°. How far
apart are the planes?
38. A ship’s captain notes that a second ship is 14.5 km away at a
bearing (see Exercise 23 of Section 9.4) of 46.3°, and that a third
ship was at a distance of 21.7 km at a bearing of 201.0°. How far
apart are the second and third ships?
39. A park is in the shape of a parallelogram with sides of 1.25 km
and 1.90 km that meet in a 78.0° angle. The park has two diagonal
paths. What is the length of each path?
9.53 mm
Fig. 9.73
34. Two ropes support a 78.3-N crate from above. The tensions in the
ropes are 50.6 N and 37.5 N. What is the angle between the
ropes? (See Exercise 28 of Section 9.5.)
35. A ferryboat travels at 11.5 km>h with respect to the water.
Because of the river current, it is travelling at 12.7 km>h with
40. A triangular machine part has sides of 5 cm and 8 cm. Explain
why the law of sines, or the law of cosines, is used to start the
solution of the triangle if the third known part is (a) the third side,
(b) the angle opposite the 8-cm side, or (c) the angle between the
5-cm and 8-cm sides.
Answers to Practice Exercises
1. B = 43.8°, A = 101.2°
2. 57.7°
CH A P T ER 9 & 26 " 5 *0/4
Vector components
y
A
Ay
uref = tan - 1
x
Ax
B
A
͉ Ay ͉
͉ Ax ͉
Law of sines
a
b
c
=
=
sin A
sin B
sin C
Law of cosines
a2 = b2 + c2 - 2bc cos A
a
b
Ay = A sin u
A = 2A2x + A2y
u
c
Ax = A cos u
b2 = a2 + c2 - 2ac cos B
C
2
2
2
c = a + b - 2ab cos C
(9.1)
(9.2)
(9.3)
(9.8)
(9.13)
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Review Exercises
CH A PTER 9
R EV IE W E XERCISES
In Exercises 1–4, find the x- and y-components of the given vectors by
use of the trigonometric functions.
1. A = 65.0, uA = 28.0°
2. A = 8050, uA = 149.0°
3. A = 0.9204, uA = 215.59°
4. A = 657.1, uA = 343.74°
In Exercises 5–8, vectors A and B are at right angles. Find the magnitude and direction of the resultant.
5. A = 327
B = 505
6. A = 6.8
B = 2.9
7. A = 4964
B = 3298
8. A = 26.52
B = 89.86
In Exercises 9–16, add the given vectors by components.
9. A = 780, uA = 28.0°
B = 346, uB = 320.0°
10. J = 0.0120, uJ = 370.5°
K = 0.00781, uK = 260.0°
11. A = 22.51, uA = 130.16°
B = 7.604, uB = 200.09°
12. A = 18 760, uA = 110.43°
B = 4835, uB = 350.20°
13. Y = 51.33, uY = 12.25°
Z = 42.61, uZ = 291.77°
14. A = 703.1, uA = 122.54°
B = 302.9, uB = 214.82°
15. A = 0.750, uA = 15.0°
B = 0.265, uB = 192.4°
C = 0.548, uC = 344.7°
16. S = 8120, uS = 141.9°
T = 1540, uT = 165.2°
U = 3470, uU = 296.0°
38. In solving a triangle for Case 3 (two sides and the included angle), explain what type of solution is obtained if the included
angle is a right angle.
39. Two angles of a triangle measure 22° and 112°. The shortest side
is 54 cm. What is the longest side?
40. Two of the angles of a triangle measure 42.0° and 59.5°. The
longest side is 5.00 cm longer than the shortest side. What is the
perimeter of the triangle?
41. An architect determines the two acute angles and one of the legs
of a right triangular wall panel. Show that the area At is
At =
a2 sin B
2 sin A
42. A surveyor determines the three angles and one side of a triangular tract of land. (a) Show that the area At can be found from
a2 sin B sin C
. (b) For a right triangle, show that this
2 sin A
agrees with the formula in Exercise 41.
At =
43. Find the horizontal and vertical components of the force shown
in Fig. 9.75.
y
17. A = 48.0°, B = 68.0°, a = 145
18. A = 132.0°, b = 7.50, C = 32.0°
19. a = 22.8, B = 33.5°, C = 125.3°
20. A = 71.0°, B = 48.5°, c = 8.42
21. A = 17.85°, B = 154.16°, c = 7863
22. a = 1.985, b = 4.189, c = 3.652
23. b = 7607, c = 4053, B = 110.09°
24. A = 77.06°, a = 12.07, c = 5.104
25. b = 14.5, c = 13.0, C = 56.6°
26. B = 40.6°, b = 7.00, c = 18.0
27. a = 186, B = 130.0°, c = 106
28. b = 750, c = 1100, A = 56°
29. a = 7.86, b = 2.45, C = 2.5°
30. a = 0.208, c = 0.697, B = 165.4°
31. A = 67.16°, B = 96.84°, c = 532.9
32. A = 43.12°, a = 7.893, b = 4.113
33. a = 17, b = 12, c = 25
34. a = 9064, b = 9953, c = 1106
y
175.6 N
In Exercises 17–36, solve the triangles with the given parts.
x
0
x
0
170.5 km/h
Fig. 9.75
Fig. 9.76
44. Find the horizontal and vertical components of the velocity
shown in Fig. 9.76.
45. In a ballistics test, a bullet was fired into a block of wood with a
velocity of 670 m>s and at an angle of 71.3° with the surface of
the block. What was the component of the velocity perpendicular to the surface?
46. A storm cloud is moving at 15 km>h from the northwest. A television tower is 60° south of east of the cloud. What is the component of the cloud’s velocity toward the tower?
47. During a 3.00-min period after taking off, the supersonic jet
Concorde travelled at 480 km>h at an angle of 24.0° above the
horizontal. What was its gain in altitude during this period?
48. A rocket is launched at an angle of 42.0° with the horizontal and
with a speed of 760 m>s. What are its horizontal and vertical
components of velocity?
49. Three forces of 3200 N, 1300 N, and 2100 N act on a bolt as
shown in Fig. 9.77. Find the resultant force.
y
36. a = 47.4, b = 40.0, c = 45.5
3200 N
1300 N 32°
In Exercises 37–71, solve the given problems.
37. For any triangle ABC show that
2
307.44°
152.48°
35. a = 0.530, b = 0.875, c = 1.25
2
293
54°
2
cos A
cos B
cos C
a + b + c
=
+
+
a
c
2abc
b
35°
Fig. 9.77
2100 N
x