polygons problems

8/13/2019 polygons problems

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514 7 ADDITIONAL TOPICS IN TRIGONOMETRY

48. Surveying. The layout in the figure is used to determine

an inaccessible height h when a baseline din a plane per-

pendicular to h can be established and the angles , , and

can be measured. Show that

h dsin csc ( ) tan

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

are given (SSS), the law of sines cannot be used to solve the triangle

neithercase 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.

(a) SAS case (b) SSS case

Law of Cosines DerivationTheorem 1 states the law of cosines.

LAW OF COSINES

a2 b2 c2 2bc cos

b2 a2 c2 2ac cos

c2 a2 b2 2ab cos

Cases SAS and SSS are most readily solved by starting with the law of

cosines.

All threeequationssayessentiallythe same

thing.

FIGURE 1


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7-2 Law of Cosines

We will establish a2 b2 c2 2bc cos . The other two equations tcan be obtained from this one simply by relabeling the figure. We start by lo

ing a triangle in a rectangular coordinate system. Figure 2 shows three typ

triangles.

For an arbitrary triangle located as in Figure 2, the distance-between-tw

points formula is used to obtain

Square both sides.

(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

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 negati


For the SAS case, start by using the law of cosines to find the side opposite

given angle. Then use either the law of cosines or the law of sines to find a s

ond angle. Because of the simpler computations, the law of sines will gener

be used to find the second angle.

h bcos

cos h

b

FIGURE 2

Three representative triangles.

h2 2hc c2 k2

a2 (h c)2 k2a

(h

c

)

2

(k

0)

2


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After using the law of cosines to find the side opposite the angle for an

SAS case, the law of sines is used to find a second angle. Figure 2

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 ormay 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

(3)

(D) Explain why equation (3) gives us the correct angle only if isacute.

The preceding discussion leads to the following strategy for solving the SAS

case:

STRATEGY FOR SOLVING THE SAS CASE

Step Find Method

1 Side opposite given angle Law of cosines

2 Second angle Law of sines

(Find the angle opposite

the shorter of the two given

sides

this angle willalways be acute.)

3 Third angle Subtract the sum of the measures

of the given angle and the angle

found in step 2 from 180.


Solve the triangle in Figure 3.

FIGURE 3

E X A M P L E

1

sin1asin b


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2

7-2 Law of Cosines

S o l u t i o n

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 uto solve for .

Solve for sin .

Solve for .

Solve for 180 ( )

180 (32.4 35.4) 112.2

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. S

sequent calculations are simplified if we solve for the obtuse angle first, if p

ent. The law of cosines is used for this purpose. A second angle, which mus

acute, can be found using either law, although computations are usually simp

with the law of sines.

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

(4

(B) Does equation (4) give us the correct angle irrespective ofwhether is acute or obtuse? Explain.

cos1a2 b2 c2

2bc

M A T C H E D P R O B L E M

1

35.4

sin16.45 sin 32.45.96

Since is acute, the inverse sine

function gives us directly. sin1

csin

b

sin csin

b

sin

c

sin

b

5.96 cm

(10.3)2 (6.45)2 2(10.3)(6.45) cos 32.4

ba2 c2 2accos

b2 a2 c2 2accos


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The preceding discussion leads to the following strategy for solving the SSS

case:

STRATEGY FOR SOLVING THE SSS CASE

Step Find Method

1 Angle opposite longest Law of cosines

side

this will take care ofan obtuse angle, if present.

2 Either of the remaining Law of sines

angles, which will be acute.

(Why?)

3 Third angle Subtract the sum of the measures

of the angles found in steps 1 and

2 from 180.


Solve the triangle with a 27.3 meters, b 17.8 meters, and c 35.2 meters.

S o l u t i o n 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.

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 .

100.5

cos1(27.3)2

(17.8)2

(35.2)2

2(27.3)(17.8)

cos1a2 b2 c2

2ab

cos a

2 b2 c2

2ab

c2 a2 b2 2abcos

FIGURE 4

E X A M P L E

2


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7-2 Law of Cosines

Solve for sin .

Solve for .

is acute.

Solve for 180

180 ( )

180 (49.7 100.5)

29.8

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 c

timeters, find the length of one side of the polygon.

S o l u t i o n Sketch a figure (Fig. 5) and use the law of cosines.

If an 11-sided regular polygon is inscribed in a circle with radius 4.63 inches, fi

the length of one side of the polygon.M A T C H E D P R O B L E M

3

19.8 centimeters

d

2(22.8)

2

2(22.8)2

cos

360

7

d2 22.82 22.82 2(22.8)(22.8) cos360

7

FIGURE 5

E X A M P L E

3

M A T C H E D P R O B L E M

2

49.7

sin127.3 sin 100.535.2

sin asin

c

27.3 sin 100.5

35.2

sin

a

sin

c


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Answers to Matched Prob lems

1. a 18.5 ft, 33.3, 69.2 2. 37.4, 95.0, 47.6 3. 2.61 inches

EXERCISE 7-2

The labeling in the figure below is the convention we will

follow in this exercise set. Your answers to some problems maydiffer slightly from those in the book, depending on the order

in which you solve for the sides and angles of a given triangle.

A

1. 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

, can you 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

B

7. Referring to the figure at the beginning of the exercise set,

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 centime-

ters. Explain why the triangle has no solution.

Solve each triangle in Problems 912 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


Problems 13

26 represent a variety of problems involvingboth the law of sines and the law of cosines. Solve each

triangle. If a problem does not have a solution, say so.

13. 94.5, 88.3, b 23.7 centimeters

14. 85.6, 97.3, a 14.3 millimeters

15. 104.5, a 17.2 inches, c 11.7 inches

16. 27.3, a 13.7 yards, c 20.1 yards

17. 57.2, 112.0, c 24.8 meters

18. 132.4, 17.3, b 67.6 kilometers

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

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,

30. Show that for any triangle,

a b cos c cos

APPLICATIONS

31. Surveying. To find the lengthAB of a small lake, a sur-

veyor measured angleACB to be 96,ACto be 91 yards,

a2 b2 c2

2abc

cos

a

cos

b

cos

c


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7-2 Law of Cosines

andBCto be 71 yards. What is the approximate length of

the lake?

32. Surveying. Suppose the figure for this problem represents

the base of a large rock outcropping on a farmers land.

If a surveyor finds ACB 110,AC 85 meters, and

BC 73 meters, what is the approximate length of the

outcropping?

33. Geometry. Find the measure in decimal degrees of a cen-

tral angle subtended by a chord of length 112 millimeters

in a circle of radius 72.8 millimeters.

34. Geometry. Find the measure in decimal degrees of a cen-

tral angle subtended by a chord of length 13.8 feet in a cir-

cle of radius 8.26 feet.

35. Geometry. Two adjacent sides of a parallelogram meet at

an angle of 3510 and have lengths of 3 and 8 feet. What

is the length of the shorter diagonal of the parallelogram

(to three significant digits)?

36. Geometry. What is the length of the longer diagonal of

the parallelogram in problem 35 (to three significant

digits)?

37. Navigation. Los Angeles and Las Vegas are approxi-

mately 200 miles apart. A pilot 80 miles from Los Angeles

finds that she is 620 off course relative to her start in Los

Angeles. How far is she from Las Vegas at this time?

(Compute the answer to three significant digits.)

38. Search and Rescue. At noon, two search planes set out

from San Francisco to find a downed plane in the ocean.

PlaneA travels due west at 400 miles per hour, and plane

B flies northwest at 500 miles per hour. At 2 P.M. planeA

spots the survivors of the downed plane and radios planeB

to come and assist in the rescue. How far is plane B from

planeA at this time (to three significant digits)?

39. Geometry. Find the perimeter of a pentagon inscribed in a

circle of radius 12.6 meters.

40. Geometry. Find the perimeter of a nine-sided regular

polygon inscribed in a circle of radius 7.09 centimeters

41. Analytic Geometry. If pointA in the figure has coordi

nates (3, 4) and pointB has coordinates (4, 3), find the

dian measure of angle to three decimal places.

42. Analytic Geometry. If pointA in the figure has coordi

nates (4, 3) and pointB has coordinates (5, 1), find the

dian measure of angle to three decimal places.

43. Engineering. Three circles of radius 2.03, 5.00, and 8.

centimeters are tangent to one another (see the figure).

Find the three angles formed by the lines joining their

ters (to the nearest 10).

44. Engineering. Three circles of radius 2.00, 5.00, and 8.

inches are tangent to each other (see the figure). Find t

three angles formed by the lines joining their centers (t

the nearest 10).

45. Geometry. A rectangular solid has sides as indicated i

the figure. Find CAB to the nearest degree.


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46. Geometry. Referring to the figure, find ACB to the

nearest degree.

47. Space Science. For communications between a space shut-

tle 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

miles above the surface of the earth (see the figure). (A

satellite ingeostationary

orbit remains stationary above afixed 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 sta-

tion to be in contact with the shuttle over most of the

Earths 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.

48. Space Science. A satellite S, in circular orbit around the

earth, is sighted by a tracking station T(see the figure).

The distance TSis 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.

Section 7-3 Geometric Vectors

Geometric Vectors and Vector Addition

Velocity VectorsForce Vectors

Resolution of Vectors into Vector Components

Many physical quantities, such as length, area, or volume, can be completely spec-

ified by a single real number. Other quantities, such as directed distances, veloc-

ities, and forces, require for their complete specification both a magnitude and a

direction. The former are often called scalar quantities, and the latter are calledvector quantities.

In this section we limit our discussion to the intuitive idea of geometric vec-

tors in a plane. In Section 7-4 we introduce algebraic vectors, a first step in the

generalization of a concept that has far-reaching consequences. Vectors are widelyused in many areas of science and engineering.

Geometric Vectors and Vector Addition

A line segment to which a direction has been assigned is called a directed line

segment. Ageometric 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 the arrowhead) is denoted by OP. Vectors are also denoted by a bold-

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