bpc5_ch07-02

516 7 Additional Topics in Trigonometry

SECTION 7-2 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 aregiven (SSS), the law of sines cannot be used to solve the triangleneither caseinvolves an angle and its opposite side (Fig. 1). Both cases can be solved startingwith the law of cosines, which is the subject matter for this section.

FIGURE 1

(a) SAS case (b) SSS case

Theorem 1 states the law of cosines.

Theorem 1 Law of Cosines

a2 b2 c2 2bc cos

b2 a2 c2 2ac cos

c2 a2 b2 2ab cos

b

a

c

Law of CosinesDerivation

ba

c

b

c

50. Surveying. The layout in the figure at right is used to de-termine an inaccessible height h when a baseline d in aplane perpendicular to h can be established and the angles, , and can be measured. Show that

h

d

h d sin csc ( ) tan

h

d

All three equationssay essentially thesame thing.

7-2 Law of Cosines 517

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 thencan be obtained from this one simply by relabeling the figure. We start by locating atriangle in a rectangular coordinate system. Figure 2 shows three typical triangles.

For an arbitrary triangle located as in Figure 2, the distance-between-two-pointsformula is used to obtain

Square both sides.

(1)

(a) (b) (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

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.]

For the SAS case, start by using the law of cosines to find the side opposite the givenangle. 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 findthe second angle.

Solving the SAS Case

h b cos

cos hb

(c, 0)

b

a

c

k

(h, 0)h 0

(h, k)

ba

c

k

h(c, 0)

(h, k)

ba

c

k

h

(h, k)

(c, 0)

FIGURE 2 Three representativetriangles.

h2 2hc c2 k2 a2 (h c)2 k2 a (hc)2 (k 0)2


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 aretwo 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 notbe obtuse. Explain why choosing the angle opposite the shorter side guaran-tees the selection of an acute angle.

(C) Starting with (sin )/a (sin )/b, show that

(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:

EXAMPLE 1 Solving the SAS Case

Solve the triangle in Figure 3.

FIGURE 3

b

10.3 cm

6.45 cm

32.4

Strategy for Solving the SAS Case

Step Find Method

1. Side opposite given angle Law of cosines

2. Second angle (Find the angle Law of sinesopposite the shorter of the two given sidesthis angle will always be acute.)

3. Third angle Subtract the sum of the measures of the given angle and the angle found in step 2 from 180.

sin1a sin b


Solution

Solve for b Use the law of cosines:

Solve for b.

Solve for Since side c is shorter than side a, must be acute, and the law of sines is used tosolve for .

Solve for sin .

Solve for .

Since is acute, the inverse sine function gives us directly.

Solve for

Matched Problem 1 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. Subse-quent calculations are simplified if we solve for the obtuse angle first, if present. Thelaw of cosines is used for this purpose. A second angle, which must be acute, can befound using either law, although computations are usually simpler with the law of sines.

EXPLORE-DISCUSS 2 (A) Starting with a2 b2 c2 2bc cos , show that

(2)

(B) Does equation (2) give us the correct angle irrespective of whether isacute or obtuse? Explain.

The above discussion leads to the following strategy for solving the SSS case.

cos1a2 b2 c2

2bc

Solving the SSS Case

180 (32.4 35.4) 112.2 180 ( )

35.4

sin16.45 sin 32.45.96

sin1c sin b

sin c sin

b

sin c

sin

b

5.96 cm

(10.3)2 (6.45)2 2(10.3)(6.45) cos 32.4 b a2 c2 2ac cos

b2 a2 c2 2ac cos


EXAMPLE 2 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 theSSS 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

Since is the largest angle, we solve for it first using the law of cosines.

Solve for

Solve for cos .

Solve for .

Solve for We now solve for either or , using the law of sines. We choose .

sin a

sin

c

100.5

cos1(27.3)2 (17.8)2 (35.2)22(27.3)(17.8)

cos1a2 b2 c2

2ab

cos a2 b2 c2

2ab

c2 a2 b2 2ab cos

17.8 m 27.3 m

35.2 m

Strategy for Solving the SSS Case

Step Find Method

1. Angle opposite longest Law of cosinessidethis will take care of an obtuse angle, if present.

2. Either of the remaining angles, Law of sineswhich will be acute (why?)

3. Third angle Subtract the sum of the measures of the angles found in steps 1 and 2from 180.


Solve for sin .

Solve for .

is acute.

Solve for

Matched Problem 2 Solve the triangle with a 1.25 yards, b 2.05 yards, and c 1.52 yards.

EXAMPLE 3 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 Sketch a figure (Fig. 5) and use the law of cosines:

FIGURE 5

Matched Problem 3 If an 11-sided regular polygon is inscribed in a circle with radius 4.63 inches, findthe length of one side of the polygon.

19.8 centimeters

d 2(22.8)2 2(22.8)2 cos 3607 d 2 22.82 22.82 2(22.8)(22.8) cos 360

7

d

Actually, youonly need tosketch thetriangle:

22.8 22.8360

7

29.8

180 (49.7 100.5) 180 ( )

180

49.7

sin127.3 sin 100.535.2

sin a sin

c

27.3 sin 100.535.2


Answers to Matched Problems

1. a 18.5 ft, 33.3, 69.22. 37.4, 95.0, 47.6 3. 2.61 in.

EXERCISE 7-2The labeling in the figure below is the convention we willfollow in this exercise set. Your answers to some problemsmay differ slightly from those in the book, depending on theorder in which you solve for the sides and angles of a giventriangle.

A1. Referring to the figure above, if 47.3, b 11.7 cen-

timeters, and c 6.04 centimeters, which of the two an-gles, 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 , canyou say for certain is acute and why?

Solve each triangle in Problems 36.

3. 71.2, b 5.32 yards, c 5.03 yards

4. 57.3, a 6.08 centimeters, c 5.25 centimeters

5. 12020, a 5.73 millimeters, b 10.2 millimeters

6. 13550, b 8.44 inches, c 20.3 inches

B7. 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 thetriangle has an obtuse angle, which angle must it be andwhy?

8. Suppose you are told that a triangle has sides a 12.5 cen-timeters, b 25.3 centimeters, and c 10.7 centimeters.Explain why the triangle has no solution.

Solve each triangle in Problems 912 if the triangle has asolution. 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 5.52kilometers

b a

c

12. a 31.5 meters, b 29.4 meters, c 33.7 meters

Problems 1326 represent a variety of problems involvingboth the law of sines and the law of cosines. Solve eachtriangle. If a problem does not have a solution, say so.13. 92.6, 88.9, a 15.2 centimeters

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


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


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

25. 39.8, a 12.5 inches, b 7.31 inches

26. 47.9, b 35.2 inches, c 25.5 inches

C27. 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,

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 22.8 sin .36014

d2

a2 b2 c2

2abc

cos

a

cos

b

cos

c


32. Show that the length d of one side of an n-sided regularpolygon, inscribed in a circle of radius r, is given byd 2r sin .

APPLICATIONS

33. Surveying. To find the length AB of a small lake, a sur-veyor measured angle ACB to be 96, AC to be 91 yards,and BC to be 71 yards. What is the approximate length ofthe lake?

34. Surveying. Suppose the figure for this problem representsthe base of a large rock outcropping on a farmers land. If asurveyor finds ACB 110, AC 85 meters, and BC 73 meters, what is the approximate length (to onedecimal place) of the outcropping?

35. Geometry. Two adjacent sides of a parallelogram meet atan angle of 3510 and have lengths of 3 and 8 feet. What isthe length of the shorter diagonal of the parallelogram (to 3significant digits)?

36. Geometry. What is the length of the longer diagonal of theparallelogram in Problem 35 (to 3 significant digits)?

37. Navigation. Los Angeles and Las Vegas are approximately200 miles apart. A pilot 80 miles from Los Angeles findsthat she is 620 off course relative to her start in Los An-geles. How far is she from Las Vegas at this time? (Com-pute the answer to 3 significant digits.)

38. Search and Rescue. At noon, two search planes set outfrom San Francisco to find a downed plane in the ocean.Plane A travels due west at 400 miles per hour, and plane Bflies northwest at 500 miles per hour. At 2 P.M. plane A spotsthe survivors of the downed plane and radios plane B tocome and assist in the rescue. How far is plane B fromplane A at this time (to 3 significant digits)?

39. Geometry. Find the perimeter of a pentagon inscribed in acircle of radius 12.6 meters.

40. Geometry. Find the perimeter of a nine-sided regular poly-gon inscribed in a circle of radius 7.09 centimeters.

A

C

B

180n

41. Analytic Geometry. If point A in the figure has coordinates(3, 4) and point B has coordinates (4, 3), find the radianmeasure of angle to three decimal places.

42. Analytic Geometry. If point A in the figure has coordinates(4, 3) and point B has coordinates (5, 1), find the radianmeasure of angle to three decimal places.

43. Engineering. Three circles of radius 2.03, 5.00, and 8.20centimeters are tangent to one another (see figure). Find thethree angles formed by the lines joining their centers (to thenearest 10).

44. Engineering. Three circles of radius 2.00, 5.00, and 8.00inches are tangent to each other (see figure). Find the threeangles formed by the lines joining their centers (to the near-est 10).

45. Geometry. A rectangular solid has sides as indicated in thefigure. Find CAB to the nearest degree.

46. Geometry. Referring to the figure, find ACB to the near-est degree.

47. Space Science. For communications between a space shut-tle and the White Sands tracking station in southern NewMexico, two satellites are placed in geostationary orbit,130 apart relative to the center of the Earth and 22,300

2.8 cm

4.3 cm8.1 cmA

B

C

x

y

A

B

0


miles above the surface of the Earth (see figure). (A satellitein geostationary orbit remains stationary above a fixedpoint on the surface of the Earth.) Radio signals are sentfrom the tracking station by way of the satellites to the shut-tle, and vice versa. This system allows the tracking stationto be in contact with the shuttle over most of the Earthssurface. How far to the nearest 100 miles is one of the geo-stationary satellites from the White Sands tracking stationW? The radius of the Earth is 3,964 miles.

S S

W

CEarth

48. Space Science. A satellite S, in circular orbit around theEarth, is sighted by a tracking station T (see figure). Thedistance TS is determined by radar to be 1,034 miles, andthe angle of elevation above the horizon is 32.4. How highis the satellite above the Earth at the time of the sighting?The radius of the Earth is 3,964 miles.

R

ST

C

Horizon

SECTION 7-3 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 speci-fied by a single real number. Other quantities, such as directed distances, velocities,and forces, require for their complete specification both a magnitude and a direc-tion. The former are often called scalar quantities, and the latter are called vectorquantities.

In this section we limit our discussion to the intuitive idea of geometric vectorsin a plane. In Section 7-4 we introduce algebraic vectors, a first step in the general-ization of a concept that has far-reaching consequences. Vectors are widely used inmany areas of science and engineering.

A line segment to which a direction has been assigned is called a directed line seg-ment. 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 withthe arrowhead) is denoted by OP. Vectors are also denoted by a boldface letter, suchas v. Since it is difficult to write boldface on paper, we suggest that you use an arrowover a single letter, such as v, when you want the letter to denote a vector.

The magnitude of the vector OP, denoted by OP, v or v, is the length of thedirected line segment. Two vectors have the same direction if they are parallel andpoint in the same direction. Two vectors have opposite direction if they are paralleland point in opposite directions. The zero vector, denoted by 0 or 0, has a magni-

Geometric Vectorsand Vector Addition

FIGURE 1 Vector OP, or v.O

v

P

bpc5_ch07-02

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