Young CAT2e SSM Ch8 Part1

543

CHAPTER 8 Section 8.1 Solutions ----------------------------------------------------------------------------------- 1. SSA 3. SSS 5. ASA 7. Observe that: ( )180 45 60 75 = + =D D D D

( )( ) ( ) 310m10m sin 60 2sin sin sin 45 sin 60 12.2m10m sin 45 2

2

ba b b

= = = = DD D

D

( )( )10m sin 75sin sin sin 45 sin 75 13.66m10m sin 45

ca c c = = =

DD DD

9. Observe that: ( )180 46 72 62 = + =D D D D ( )( )200cm sin 46sin sin sin 46 sin 62 163cm

200 cm sin 62a

a b a = = =

DD DD

( )( )200cm sin 72sin sin sin 62 sin 72 215cm200cm sin 62

cb c c = = =

DD DD

11. Observe that: ( )180 16.3 47.6 116.1 = + =D D D D ( )( )211 yd. sin16.3sin sin sin16.3 sin 47.6 80.2 yd.

211 yd. sin 47.6a

a c a = = =

DD DD

( )( )211 yd. sin116.1sin sin sin116.1 sin 47.6 256.6 yd.211 yd. sin 47.6

bb c b = = =

DD DD

13. Observe that: ( )180 30 30 120 = + =D D D D ( )( )12m sin 30sin sin sin 30 sin120 7 m

12 m sin120a

a c a = = =

DD DD

( )( )12m sin 30sin sin sin 30 sin120 7 m12m sin120

bb c b = = =

DD DD

Chapter 8

544

15. Observe that: ( )180 26 57 97 = + =D D D D ( )( )100 yd. sin 97sin sin sin 97 sin 57 118 yd.

100 yd. sin 57a

a c a = = =

DD DD

( )( )100 yd. sin 26sin sin sin 26 sin 57 52 yd.100 yd. sin 57

bb c b = = =

DD DD

17. Step 1: Determine . 1sin sin sin16 sin 5sin16 5sin16sin sin 20

4 5 4 4a b = = = =

D D D D

This is the solution in QI label it as 1 . The second solution in QII is given by 2 1180 160 = D D . Both are tenable solutions, so we need to solve for two triangles.

Step 2: Solve for both triangles. QI triangle: 1 20 D ( )1 180 16 20 144 + =D D D D 1 1

1 1

sinsin sin16 sin144 4sin144 94 sin16

ca c c

= = = D D D

D

QII triangle: 2 160 D ( )2 180 16 160 4 + =D D D D 2 2

2 2

sinsin sin16 sin 4 4sin 4 14 sin16

ca c c

= = = D D D

D

19. Step 1: Determine . sin sin sin 40 sin sin sin 40 40

12 12c a = = = =

D D D

Note that there is only one triangle in this case since the angle in QII with the same sine as this value of is 180 40 140 =D D D . In such case, note that 180 + = D , therefore preventing the formation of a triangle (since the three interior angles, two of which are and , must sum to 180D - this would only occur if the third angle 0 = , in which case there is no triangle). Step 2: Solve for the triangle. ( )180 40 40 100 + =D D D D sin sin sin 40 sin100 12sin100 18

12 sin 40b

c b b = = =

D D DD

Section 8.1

545

21. Step 1: Determine . 1sin sin sin sin100 21sin100 21sin100sin sin

21 14 14 14a b = = = =

D D D,

which does not exist. Hence, there is no triangle in this case. 23. Step 1: Determine .

( ) ( )12 118sin sin sin 30 sin sin 1 sin 1 909 18 9a b

= = = = = = DD

.

Step 2: Solve the triangle. ( )180 90 30 60 + =DD D D sin 30 sin 60 9sin 60 9 3 16

9 sin 30c

c= = =

D D D

D

25. First, note that there is only one triangle since a b and 0 sin 1< < . Step 1: Determine .

1sin sin 7sin 34 7sin 34sin sin 2310 10a b

= = = D D D

Step 2: Solve for the triangle. ( )180 34 23 123 + =DD D D sin sin sin 34 sin123 10sin123 15

10 sin 34c

a c c = = =

DD DD

Chapter 8

546

27. First, note that we expect two triangles since is acute and a < b. Step 1: Determine .

1sin sin sin 21.3 sin 6.18sin 21.3 6.18sin 21.3sin sin 21.96.03 6.18 6.03 6.03a b

= = = = D D D D

This is the solution in QI label it as 1 . The second solution in QII is given by 2 1180 158.1 = D D . Both are tenable solutions, so we need to solve for two triangles.

Step 2: Solve for both triangles. QI triangle: 1 21.9 D ( )1 180 21.3 21.9 136.8 + =D D D D 1 1

1 1

sinsin sin 21.3 sin136.8 6.03sin136.8 11.366.03 sin 21.3

ca c c

= = = D D

D

QII triangle: 2 158.1 D ( )2 180 21.3 158.1 0.6 + =D D D D 2 2

2 2

sinsin sin 21.3 sin 0.6 6.03sin 0.6 0.176.03 sin 21.3

ca c c

= = = D D D

D

29. First, note that we expect only one triangle since is obtuse and a > b. Step 1: Determine . sin sin 4 3 sin116sin 0.880635 62

5 2a b = =

D D

Step 2: Solve for the triangle. ( )180 116 62 2 + =DD D D sin sin sin116 sin 2 5 2 sin 38 0.275

sin1165 2c

a c c = = =

DD DD

Section 8.1

547

31. Step 1: Determine . 1sin sin sin 40 sin 500sin 40 500sin 40sin sin 77

330 500 330 330c b = = = =

D D D D

This is the solution in QI label it as 1 . The second solution in QII is given by 2 1180 103 = D D . Both are tenable solutions, so we need to solve for two triangles.

Step 2: Solve for both triangles. QI triangle: 1 77 D ( )1 180 77 40 63 + =D D D D 1

11 1

sin sin sin 63 sin 40 330sin 63 457330 sin 40

aa c a = = =

D D DD

QII triangle: 2 103 D ( )2 180 103 40 37 + =D D D D 2 2

2 2

sin sin sin 37 sin 40 330sin 37 309330 sin 40

aa c a = = =

D D DD

33. Step 1: Determine . 1sin sin sin sin106 2 sin106 2 sin106sin sin 31

2 7 7 7a b = = = =

D D D D

Note that there is only one triangle in this case since the angle in QII with the same sine as this value of is 180 31 149 =D D D . In such case, note that 180 + > D , therefore preventing the formation of a triangle (since the three interior angles, two of which are and , must sum to180D ). Step 2: Solve for the triangle. ( )180 31 106 43 + =D D D D sin sin sin 43 sin106 7 sin 43 2

sin1067c

c b c = = =

D D DD

Chapter 8

548

35. Consider the following diagram:

First, using the properties of corresponding and supplementary angles, we see that

80 = D . As such, the information provided yields an AAS triangle. So, we use Law of Sines to solve the triangle: ( )180 9 80 91 = + =D D D D

( )sin 9 sin 91195

195 sin 911, 246 ft.

sin 9

w

w

=

=

D D

DD


We need to determine H. To this end, first observe that 154.5 = D , and so ( )180 20.5 154.5 5 = + =D D D D . So, the information provided yields an AAS situation for triangle T1. So, use Law of Sines to find x: sin 20.5 sin 5 sin 20.5 4.01818mi.

1 mi. sin 5x

x= = D D D

D

Now, to find H, we turn our focus to triangle T2. Again, from the information we have, the situation is AAS, so we use Law of Sines to find H:

sin 90 sin 25.5 1 sin 25.54.01818x H H

= =D D D

so that ( )4.01818 sin 25.5 1.7 mi.H D

9D

25.5D20.5D

Section 8.1

549


We seek x. To this end, first note that 50 = D . So, the information provided yields an AAS situation for triangle T1. So, we use Law of Sines to find x: ( )1 mi. sin100sin100 sin 50 1.3mi.

1 mi. sin 50x

x= =

DD DD


We seek x. The information provided yields an AAS situation for triangle T1. So, we use Law of Sines to find x:

( )100 ft. sin15sin15 sin100 26 ft.100 ft. sin100

xx

= = DD D

D

30D100D

15D65D

100D

Chapter 8

550


Using the Law of Sines yields sin 41 sin104

400400sin 41 270 ft.

sin104

PR

PR

=

=

D D

D

D


Using the Law of Sines yields

1

sin sin 56 20.3sin 56sin20.3 19.4 19.4

20.3sin 56sin 60.16919.4

= =

=

D D

DD

So,

( )180 56 60.169 63.83 = + =D D D D

41D35D

104D

56D

Section 8.1

551


They travel the same distance since the triangle is isosceles. To find this distance, first note that ( )180 2 84 12 = =D D D . So, using the Law of Sines yields

sin12 sin8416

16sin84 76.5 ft.sin12

A

A

=

=

D D

D

D


First, note that ( )180 72 16 92 = + =D D D D . Using the Law of Sines yields

sin 92 sin 7263.2

63.2sin 72 60.14 ft.sin 92

A

A

=

=

D D

D

D

51. The value of is incorrect. Should be 9sin120sin7

=D

.

53. False. It applies to all triangles. 55. True. By definition of oblique. 57. True

84D84D

16D

72D

Chapter 8

552

59. Claim: ( ) ( )1sin cos2 2

a b c + = Proof: First, note that the given equality is equivalent to:

( ) ( )1cos 2sin

2

a bc

+ =

We shall start with the right-side, and we will need the following identities:

sin sin 2sin cos ( )2 2

sin sin 2 2sin cos2 2 2

+ + = = =

1

(2)

Indeed, observe that sin sin sin sinsin sin sin

2sin cos2 2 (by )

sin

2sin cos2 2 (by )

2sin cos2 2

a b a bc c c

+ += + = + =+ =+ =

(1)

(2)

Now, since + + = , it follows that 2 2 2

+ = . As such, sin sin cos

2 2 2 2 + = = (3).

Using (3) in the last line of the above string of equalities then yields

22sin cos2 2

2sin cos2 2

+ =

cos2 cos 2

2

sin cos

2 2

cos2

sin2

= ,

as desired. 61. Observe that since all sides have the same lengths, the Law of Sines yields

sin sin sina a a = = ,

from which we immediately conclude that = = since the angles must sum to 180D and none of them can be obtuse.

Section 8.1

553

63. The following steps constitute the program in this case: Program: AYZ :Input SIDE A =, A :Input ANGLE Y =, Y :Input ANGLE Z =, Z :180-Y-Z X :Asin(Y)/sin(X)B :Asin(Z)/sin(X)C :Disp ANGLE X = , X :Disp SIDE B = , B :Disp SIDE C =, C Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 10 ANGLE Y = 45 ANGLE Z = 65 Now, the program will display the following information that solves the triangle: ANGLE X = 70 SIDE B = 7.524873193 SIDE C = 9.64472602

65. The following steps constitute the program in this case: Program: ABX :Input SIDE A =, A :Input SIDE B =, B :Input ANGLE X =, X :sin-1(Bsin(X)/A)Y :180-Y-X Z :Asin(Z)/sin(X)C :Disp ANGLE Y = , Y :Disp ANGLE Z = , Z :Disp SIDE C =, C Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 22 SIDE B = 17 ANGLE X = 105 Now, the program will display the following information that solves the triangle: ANGLE Y = 48.27925113 ANGLE Z = 26.72074887 SIDE C = 10.24109154

Chapter 8

554

67. The following steps constitute the program in this case: Program: ACX :Input SIDE A =, A :Input SIDE C =, C :Input ANGLE X =, X :sin-1(Csin(X)/A)Y :180-Y-X Z :Asin(Z)/sin(X)B :Disp ANGLE Y = , Y :Disp ANGLE Z = , Z :Disp SIDE B =, B Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 25.7 SIDE C = 12.2 ANGLE X = 65 Now, the program will display the following information that solves the triangle: ANGLE Y = 25.48227 ANGLE Z = 89.51773 SIDE C = 28.35581

Section 8.2 Solutions ------------------------------------------------------------------------------------ 1. C 3. S 5. S 7. C 9. This is SAS, so begin by using Law of Cosines: Step 1: Find b.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 4 3 2 4 3 cos 100 25 24cos 1005.4 5

b a c acb

= + = + =

D D

Step 2: Find . 1sin sin sin100 sin 3sin100 3sin100sin sin 33

5 3 5 5b c = = = =

D D D D

Step 3: Find : ( )180 33 100 47 + =D D D D

Section 8.2

555

11. This is SAS, so begin by using Law of Cosines: Step 1: Find a.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 7 2 2 7 2 cos 16 53 28cos 165.107 5

a b c bca

= + = + =

D D

Step 2: Find . 1sin sin sin16 sin 2sin16 2sin16sin sin 6

5.107 2 5.107 5.107a c = = = =

D D D D

Step 3: Find . ( )180 6 16 158 + =D D D D 13. This is SAS, so begin by using Law of Cosines: Step 1: Find a.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 5 5 2 5 5 cos 20 50 50cos 201.736 2

a b c bca

= + = + =

D D

Step 2: Find . 1sin sin sin 20 sin 5sin 20 5sin 20sin sin 80

1.736 5 1.736 1.736a b = = = =

D D D D

Step 3: Find . ( )180 80 20 80 + =D D D D 15. This is SAS, so begin by using Law of Cosines: Step 1: Find b.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 9 12 2 9 12 cos 23 225 216cos 235.1158 5

b a c acb

= + = + =

D D


5.1158 9 5.1158 5.1158b a = = = =

D D D D

Step 3: Find . ( )180 43 23 114 + =D D D D

Chapter 8

556

17. This is SAS, so begin by using Law of Cosines: Step 1: Find b.

( ) ( ) ( )( ) ( ) ( )2 22 2 2 2 cos 4 8 2 4 8 cos 60 80 64cos 606.9282 7

b a c acb

= + = + =

D D


6.9282 4 6.9282 6.9282b a = = = =

D D D D

Step 3: Find . ( )180 30 60 90 + =D D D D 19. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is .

( ) ( ) ( ) ( ) ( )2 2 22 2 2 2 cos 8 5 6 2 5 6 cos 64 61 60cosa b c bc = + = + = Thus, 13 3cos so that cos 92.86 93

60 60 = =

D D .

Step 2: Find either of the remaining two angles using the Law of Sines. 1sin sin sin 92.866 sin 5sin 92.866 5sin 92.866sin sin 39

8 5 8 8a b = = = =

D D D D

Step 3: Find the third angle. ( )180 93 39 48 + =D D D D 21. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is .

( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 5 4 4 2 4 4 cos 25 32 32cosc a b ab = + = + = Thus, 17 7cos so that cos 77

32 32 = =

D .

Step 2: Find either of the remaining two angles using the Law of Sines. 1sin sin sin 77.364 sin 4sin 77.364 4sin 77.364sin sin 51.32

5 4 5 5c b = = = =

D D D D

Step 3: Find the third angle. ( )180 77.36 51.32 51 + D D D D

Section 8.2

557

23. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is .

( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 8.2 7.1 6.3 2 7.1 6.3 cos 67.24 90.1 89.46cosa b c bc = + = + =

Thus, 122.86 22.86cos so that cos 75.19 7589.46 89.46

= = D D .

Step 2: Find either of the remaining two angles using the Law of Sines. 1sin sin sin 75.19 sin 7.1sin 75.19 7.1sin 75.19sin sin 57

8.2 7.1 8.2 8.2a b = = = =

D D D D

Step 3: Find the third angle. ( )180 75 57 48 + =D D D D 25. Since 9a b+ = >10 c= , there can be no triangle in this case. 27. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is .

( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 13 12 5 2 12 5 cos 0 120cosc a b ab = + = + = Thus, ( )1cos 0 so that cos 0 90 = = = D . Step 2: Find either of the remaining two angles using the Law of Sines.

1sin sin sin 90 sin 5sin 90 5sin 90sin sin 2313 5 13 13c b

= = = = D D D D

Step 3: Find the third angle. ( )180 90 23 67 + =D D D 29. This is AAS, so begin by using Law of Sines: ( )180 40 35 105 = + =D D D D

( )( )6 sin 35sin sin sin 35 sin 40 5.354 56 sin 40

ba b b = = =

DD DD

( )( )6 sin105sin sin sin 40 sin105 96 sin 40

ca c c = = =

DD DD

Chapter 8

558

31. This is SSA, so begin by using Law of Sines: Step 1: Determine .

1sin sin sin 31 sin 5sin 31 5sin 31sin sin 12.392 1212 5 12 12a b

= = = = D D D D D

Note that there is only one triangle in this case since the angle in QII with the same sine as this value of is 180 12 168 =D D D . In such case, note that 180 + > D , therefore preventing the formation of a triangle (since the three interior angles, two of which are and , must sum to180D ). Step 2: Solve for the triangle. ( )180 12 31 137 + =D D D D sin sin sin137 sin 31 12sin137 16

12 sin 31c

c a c = = =

D D DD

33. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is .

( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 8 7 3 2 7 3 cos2 2 21cos

b a c ac

= + = + =

Thus, 11 1cos so that cos 77.396 7721 21

= = D D .

Step 2: Find either of the remaining two angles using the Law of Sines. 1sin sin sin sin 77.396 7 sin 77.396 7 sin 77.396sin sin 66

7 8 8 8a b = = = =

D D D D

Step 3: Find the third angle. ( )180 77 66 37 + =D D D D

Section 8.2

559

35. This is SSA, so begin by using Law of Sines: Step 1: Determine .

1sin sin sin sin10 2sin10 2sin10sin sin 1.809 22 11 11 11c b

= = = = D D D D D

Note that there is only one triangle in this case since the angle in QII with the same sine as this value of is 180 2 178 =D D D . In such case, note that 180 + > D , therefore preventing the formation of a triangle (since the three interior angles, two of which are and , must sum to 180D ). Step 2: Solve for the triangle. ( )180 2 10 168 + =D D D D sin sin sin168 sin10 11sin168 13

11 sin10a

a b a = = =

D D DD

37. This is SAS, so begin by using Law of Cosines: Step 1: Find b.

( ) ( ) ( )( ) ( )2 22 2 2 48 71 48 715 5 5 52 cos 2 cos 51.62 124.524863111.16

b a c acb

= + = +

D

Step 2: Find . 71 71

15 5

715

sin sin sin 51.62 sin sin 51.62 sin 51.62sin sin 85.9111.156 11.156 11.156b c

= = = = D D D D

Step 3: Find . ( )180 85.91 51.62 40.40 + =D D D D 39. This is SSS, so begin by using Law of Cosines: Step 1: Find the largest angle (i.e., the one opposite the longest side). Here, it is .

( ) ( ) ( ) ( )( )2 2 22 2 2 2 cos 24.12 18.21 21.30 2 18.21 21.30 cos203.5197 775.746cos

c a b ab

= + = + =

Thus, ( )1cos 0.2623535 so that cos 0.2623535 74.79 = D . Step 2: Find either of the remaining two angles using the Law of Sines.

1sin sin sin sin 74.79 18.21sin 74.79 18.21sin 74.79sin sin 46.7618.21 24.12 24.12 24.12a c

= = = = D D D D

Step 3: Find the third angle. ( )180 74.79 46.76 58.45 + =D D D D

Chapter 8

560

41. This is SAS, so begin by using Law of Cosines: Step 1: Find a.

( ) ( ) ( )( ) ( )2 22 2 2 5 2 3 2 5 2 3 27 5 7 52 cos 2 cos 71.213 1.1883209081.090

a b c bca

= + = +

D

Step 2: Find . 5 2 5 2

17 75 2

7

sin 71.213 sin 71.213sin sin sin sin 71.213 sin sin 611.090 1.090 1.090b a

= = = = D DD D

Step 3: Find . ( )180 71.213 61.33 47 + D D D D 43. Consider the following diagram:

Let x = distance between Plane 1 and Plane 2 after 3 hours. In order to determine the value of x, we apply the Law of Cosines as follows:

2 2 21500 1305 2(1500)(1305)cos1507,343,514.456 2710 mi.

xx= + =

D

Hence, the distance between the two planes after 3 hours is approximately 2,710 mi. .

45. Let P1 be the position of plane 1, P2 be the position of plane 2, and 65 = D the angle between OP1 and OP2 (where O is the origin). Since distance = rate * time, the length of OP1 is 1375 miles and the length of OP2 is 875 miles. Using the Law of Cosines yields

( )2 2 21 21 2

1375 875 2(1375)(875)cos651280

PPPP

= +

D

So, the planes are approximately 1,280 miles apart after 2.5 hours.

150D

Section 8.2

561


Since the segment connecting the Pitchers mound to Home Base lies on the diagonal of the square (connecting home base to second base), we know that 45 = D (since diagonals bisect the angles made at their respective vertices). We use the Law of Cosines to find x:

2 2 290 60.5 2(90)(60.5)cos 45 so that 63.7 ft.x x= + D So, the Pitchers mound is approximately 63.7 ft. from Third Base. 49. Consider the following diagram:

Let x = length of the window. First, observe that ( )180 40 125 15 + =D D D D . So, sin15 sin 40 40sin15 16 ft.

40 ft. sin 40x

x= = D D D

D .

42.2D

40D

125D

55D

Chapter 8

562


Let x = distance from Pegs house to school. Then, we obtain

( )1 mi. sin 50sin 50 sin 70 0.8 mi.1 mi. sin 70

xx

= = DD D

D

53. Consider the following diagram: First, note that the right triangle at the top of the diagram has legs of lengths 30 feet and 40 feet. Hence, by the Pythagorean theorem, the zipline has length 50 feet. Next, find b using the Law of Cosines:

2 2 250 120 2(50)(120)cos5095.85

bb= +

D

Now, use the Law of Sines to find : 1sin sin 50 50sin 50sin

50 95.85 95.85

21.67

= =

D D

D

60D 50D

70D

Section 8.2

563


First, using the Law of Cosines yields 2 2 27 1 2(7)(1)cos 25

6.108 ft.QRQR

= +

D

Hence, using the Law of Sines, we obtain

1sin sin 25 7sin 25sin 151.037 6.108 6.108 = =

D DD

Since the angles of a triangle must sum to 180D , we see that 3.97 = D .


First, by the Law of Cosines, we see that

( ) ( )22 2 7 712 1263 62.8 2(62.8) cos 70.1987 = + D . So, by the Law of Sines, we have

712

sin sin 70.1987 5063

= D

D

25D

Chapter 8

564


Using the Law of Cosines yields 2 2 2650 480 500 2(480)(500)cos

83.07

= + D

61. Should have used the smaller angle in Step 2. 63. False. Can use Law of Cosines to solve any such triangle. 65. True. By the Law of Cosines,

2 2 2 2 2

0

2 cos90c a b ab a b=

= + = +D So, the Pythagorean Theorem is a special case of the Law of Cosines.

67. True. Each of the other two angles are 180

2D

. If you have the length of the side adjacent to the angle (and hence opposite one of the two congruent sides), then you have the other side as well since the triangle is isosceles. Hence, by SAS, you can apply the Law of Cosines.

Section 8.2

565

69. Claim: 2 2 2cos cos cos

2a b c

a b c abc + ++ + =

Proof: First, observe that the left-side simplifies to: cos cos cos cos cos cosbc ac ab

a b c abc + ++ + = (1).

Next, note that the Law of Cosines yields the following three identities: ( ) ( ) ( )2 2 2 2 2 2 2 2 2cos , cos , cos

2 2 2a b c b a c c a b

bc ac ab + + += = = Now, substituting these into the right-side of (1) yields:

( ) ( ) ( )2 2 2 2 2 2 2 2 22 2 2 2 2 2

cos cos cos cos cos cos

2

2 2

bc ac aba b c abc

a b c b a c c a babc

a b c a b cabc abc

+ ++ + = + + = + + + += =

Chapter 8

566


Observe that since 2 180X + = D , we know that 1802 X = D . By the Law of Cosines, we have

( ) ( )( )

222 2 22 2 4

11 14 4

2 cos cos

cos cos

a a aa a a a

= + = = =

Hence, ( ) ( )1 11801 14 2 4cos 180 2cosX X = = D D .

So, we now have

( ) ( )( ) ( )( )1 11 14 42 1 cos 180 2cos 1 cos 2cos1 coscos 2 2 2X X + += = =

D

Section 8.2

567

73. The following steps constitute the program in this case: Program: BCX :Input SIDE B =, B :Input SIDE C =, C :Input ANGLE X =, X : (B2+C2-2BCcos(X))A :sin-1(Bsin(X)/A)Y :180-Y-X Z :Disp SIDE A = , A :Disp ANGLE Y = , Y :Disp ANGLE Z =, Z Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE B = 45 SIDE C = 57 ANGLE X = 43 Now, the program will display the following information that solves the triangle: SIDE A = 39.01481143 ANGLE Y = 51.87098421 ANGLE Z = 85.12901579

75. The following steps constitute the program: Program: ABC :Input SIDE A =, A :Input SIDE B =, B :Input SIDE C =, C :cos-1((B2+C2-A2)/(2BC)) X : cos-1((A2+C2-B2)/(2AC)) Y :180-Y-X Z :Disp ANGLE X = , X :Disp ANGLE Y = , Y :Disp ANGLE Z =, Z Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 29.8 SIDE B = 37.6 SIDE C = 53.2 Now, the program will display the following information that solves the triangle: ANGLE X = 32.98051035 ANGLE Y = 43.38013581 ANGLE Z = 103.6393538

77. The following steps constitute the program: Program: ABX :Input SIDE A =, A :Input SIDE B =, B :Input ANGLE X =, X : (A2+B2-2ABcos(X))C :sin-1(Asin(X)/C)Y :180-Y-X Z :Disp SIDE C = , C :Disp ANGLE Y = , Y :Disp ANGLE Z =, Z Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 12 SIDE B = 21 ANGLE X = 43 Now, the program will display the following information that solves the triangle: SIDE C = ANGLE Y = ANGLE Z =

Chapter 8

568

Section 8.3 Solutions ------------------------------------------------------------------------------------

1. ( )( )1 8 16 sin 60 55.42

A = D 3. ( )( )1 1 2 sin 45 0.52A = D 5. ( )( )1 6 8 sin80 23.6

2A = D 7. ( )( )1 4 7 sin 27 6.4

2A = D

9. ( )( )1 100 150 sin 36 4,408.42

A = D 11. ( )( )1 5 5 5 sin 50 9.62A = D 13. First, observe that 45 22.5

2 2a b cs + += = = . So, the area is given by

( )( )( ) ( )( )( )( )22.5 7.5 7.5 7.5 97.4A s s a s b s c= = 15. First, observe that 17 51

2 2a b cs + + += = . So, the area is given by

( )( )( )17 51 17 51 17 51 17 517 51 10 25.0

2 2 2 2

A s s a s b s c= + + + +=

17. First, observe that 25 12.52 2

a b cs + += = = . So, the area is given by ( )( )( ) ( )( )( )( )12.5 6.5 2.5 3.5 26.7A s s a s b s c= =

19. First, observe that 25.052

a b cs + += = . So, the area is given by ( )( )( ) ( )( )( )( )25.05 10.75 9.35 4.95 111.64A s s a s b s c= =

21. First, observe that 24,6002

a b cs + += = . So, the area is given by ( )( )( ) ( )( )( )( )24,600 10,600 8,100 5,900 111,632,076A s s a s b s c= =

23. Observe that the sum of the lengths of the shortest two sides is 155, which does not exceed the length of the third side. Hence, there is no triangle in this case.

25. ( )( )1 2 3 5 3 sin 61.23 13.152A = D 27. ( )( )86 513 41 sin 73 174.762A = D

Section 8.3

569

29. First, observe that 5.61 6.61 4.312 2

a b cs + + + += = . So, the area is given by ( )( )( )5.61 6.61 4.31 5.61 6.61 4.31 5.61 6.61 4.31 5.61 6.61 4.315.61 6.61 4.31

2 2 2 211.98

A s s a s b s c= + + + + + + + + =

31. First, observe that 63 50 508 7 9

2 2a b cs + ++ += = . So, the area is given by

( )( )( )63 50 50 63 50 50 63 50 50 63 50 508 7 9 8 7 9 8 7 9 8 7 963 50 50 19.21

2 2 8 2 7 2 9

A s s a s b s c= + + + + + + + + =

33. First, observe that 1, 2772

a b cs + += = . So, the area is given by ( )( )( ) ( )( )( )( )1, 277 427 467 383 312, 297 sq. nm.A s s a s b s c= =

35. First, observe that 237.52

a b cs + += = . So, the area is given by ( )( )( ) ( )( )( )( )237.5 77.5 97.5 62.5 10,591 sq. ft.A s s a s b s c= =

37. We need to determine such that ( )( )11000 60 120 sin2

= . This is equivalent to 1000sin3600

= so that 1 1000sin 163600

= D . Since there are TWO triangles in this case,

we note that the other one yields 164 = D . 39. First, observe that 44.6

2a b cs + += = . So, the area is given by

( )( )( ) ( )( )( )( )44.6 22.1 16.5 6 312.4 sq. mi.A s s a s b s c= =

Chapter 8

570

41. Consider an equilateral triangle T each of whose sides has a length of x. We wish to

determine x such that 2

29Area(T) in.2

= . To this end, observe that 3

2 2x x xs x+ += = . So, the area is given by

( )( )( ) 3 3 4 23 3 3 1 3 32 2 2 2 16 4

A s s a s b s c x x x x x x x = = = = = .

So, we need to solve the following equation: 23 814 4

x =

Indeed, observe that 2 81 44 3

x = so that 12 in.x 43. Let T be a triangle with sides 200 ft., 260 ft., and included angle 65D . The area of the parallelogram is equal to 2 area of T. Note that area of T = ( )( )1 200 260 sin 65 23,564 sq. ft.

2D

So, the area of the parallelogram is approximately 47,128 sq. ft. 45. First, note that a regular hexagon can be dissected into 6 congruent equilateral triangles

this is the case since ( )180 6 2 1206

= =D

D , and diagonals bisect the interior angles,

thereby making two of the three angles in each of the triangles equal to 60D . As such, area of a regular hexagon = 6 area of one these triangles. To find the area of one these triangles, we apply Herons formula.

Observe that 3 3 3 4.52

s + += = and so, area ( )34.5 4.5 3 3.897 sq. ft.A = = Thus, the area of the regular hexagon is ( )6 3.897 sq. ft. 23.38 sq. ft.= .

Section 8.3

571


We compute the area of the three disjoint triangles separately:

12 1 2

13 2 2

12 3 1 2

Area of (6)(8)sin 30



Q PQ

Q Q P

Q Q Q

===

D

D

D

+++

Summing these areas yields the area of the outer triangle 1 3PQ Q+ as approximately 49.77 square units. This does not equal the area of the outer triangle, which is about 35.5 sq. units. The reason these arent equal is because the outer triangle cannot exist as drawn. .


First, note that 12Area of (2)(3)sin135 2.1213ABC = D+ .

Next, in order to find R, we apply the Law of Cosines:

2 2 22 3 2(2)(3)cos135 21.485284.63522

RR= +

D

Now, in order to find the area of ACD , observe that

5 7 4.63522 8.317612 2

a b cs + + + += = . Thus, the area of ACD is given by

( )( )( )4 5 4.63522 8.73A s s s s= . Hence, the area of the entire quadrilateral is approximately 10.86 square units.

51. The semiperimeter is half the perimeter, namely 2

a b cs + += . 53. True. The formula works for all triangles. 55. True. This follows directly from Henons formula for area of a triangle.

135D

Chapter 8

572

57. False. A square with side lengths 2 units has an area of 4 square units, while a rectangle with adjacent sides having lengths 1 unit and 4 units, also has an area of 4 square units. However, none of the corresponding sides are congruent between this particular square and rectangle. 59. Consider the following diagram:

Claim: Area = 2 sin sin

2sina

. Proof: First, the Law of Sines gives yields sin sin sin

sinax

a x

= = .

Hence, the area = ( ) 21 sin sin sinsin2 sin 2sin

a aa = .


Observe that Area of sector AOB =

( ) ( )( )212 1805 40 8.72664626 DD and

Area of AOB = ( )( )12 5 5 sin 40 8.034845D .

Thus, the area of the shaded region is Area of sector AOB - Area of AOB

0.69 square units.

Section 8.3

573


Draw in the dashed diagonal above. Note that the triangles formed are congruent, and each has an area of 2 21 12 4sin 30s s=D .The area of the rhombus, therefore, must be ( )2 21 14 22 s s= . 65. The following steps constitute the program in this case: Program: ABZ :Input SIDE A =, A :Input SIDE B =, B :Input ANGLE Z =, Z :1/2*ABsin(Z)W :Disp AREA W = , W Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 35 SIDE B = 47 ANGLE Z = 68 Now, the program will display the following information that gives the area of the triangle: AREA W = 762.6087204

67. The following steps constitute the program in this case: Program: ABC :Input SIDE A =, A :Input SIDE B =, B :Input SIDE C =, C :(A+B+C)/2S : (S(S-A)(S-B)(S-C))W :Disp AREA W = , W Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 85 SIDE B = 92 SIDE C = 123 Now, the program will display the following information that gives the area of the triangle: AREA W = 3907.492802

30D

Chapter 8

574

69. The following steps constitute the program in this case: Program: ABC :Input SIDE A =, A :Input SIDE B =, B :Input SIDE C =, C :(A+B+C)/2S : (S(S-A)(S-B)(S-C))W :Disp AREA W = , W Now, in order to use this program to solve the given triangle, EXECUTE it and enter the following data at the prompts: SIDE A = 145 SIDE B = 172 SIDE C = 110 Now, the program will display the following information that gives the area of the triangle: AREA W =

Section 8.4 Solutions ------------------------------------------------------------------------------------ 1.

( ) ( )2 25 2,9 7 3,2 3 2 13AB AB= = = + =JJJG JJJG 3.

( ) ( )2 23 4,0 1 7, 1 7 1 50 5 2AB AB= = = + = =JJJG JJJG 5.

( ) ( )2 224 0,0 7 24, 7 24 7 25AB AB= = = + =JJJG JJJG 7. Given that 3,8u =G , we have ( ) ( )2 23 8 73u = + =G and 8tan

3 = so that

1 8tan 69.43

= D .

Section 8.4

575

9. Given that 5, 1u = G , we have ( ) ( )2 25 1 26u = + =G and 1tan5

= so that 1 1tan 348.7

5 =

D .

11. Given that 4,1u = G , we have ( ) ( )2 24 1 17u = + =G and 1tan4

= so that since

the head is in QII, 1 1180 tan 166.04

= + D D .

13. Given that 8,0u = G , we have ( ) ( )2 28 0 8u = + =G . Since the vector is on the x-axis pointing in the negative direction, 180 = D . 15. Given that 3,3u =G , we have ( ) ( )2 23 3 2 3u = + =G and 3tan

3 = so that

1 3tan 603

= D .

17. 4,3 2, 5 4 2,3 5 2, 2u v+ = + = + = G G . 19. 3 3 4,3 12,9u = = G 21. 2 4 2 4,3 4 2, 5 8,6 8, 20 0, 14u v+ = + = + = G G 23. ( )6 6 6 6 4,3 6 2, 5 24,18 12,30 36,48u v u v = = = + = G G G G 25. cos , sin 7cos 25 , 7sin 25 6.3, 3.0u u u = = D DG G G 27. cos , sin 16cos100 , 16sin100 2.8, 15.8u u u = = D DG G G 29. cos , sin 4cos310 , 4sin 310 2.6, 3.1u u u = = D DG G G 31. ( ) ( )cos , sin 9cos 335 , 9sin 335 8.2, 3.8u u u = = D DG G G 33. cos , sin 2cos120 , 2sin120 1, 1.7u u u = = D DG G G

35. ( ) ( )2 25, 12 5, 12 5 12,

13 13 135 12

vv

= = = +

GG

37. ( ) ( )2 260,11 60,11 60 11,

61 61 6160 11

vv= = =

+GG

Chapter 8

576

39. ( ) ( )2 224, 7 24, 7 24 7,

25 25 2524 7

vv

= = =+

GG

41. ( ) ( )2 29, 12 9, 12 9 12 3 4, ,

15 15 15 5 59 12

vv

= = = = +

GG

43. ( ) ( )2 22,3 2 2,3 2 2 3 2 10 3 10, ,

10 102 5 2 5 2 52 3 2

vv= = = =

+

GG

45. 7 3i j+G G 47. 5 3i jG G 49. 0i j +G G 51. 2 0i j+G G 53. 5 5i j +G G 55. 7 0i j+G G 57. The velocity vector is given by 2200cos30 , 2200sin 30 1905, 1100v = D DG . So, the horizontal component is approximately ft.1,905 sec. and the vertical component is

approximately ft.1,100 sec. .

59. weight = 630 lbs. 2800.609 lbs. 2801 lbs.sin13

D

Section 8.4

577


By the Pythagorean Theorem, the actual velocity of the ship is

2 210 6 11.6619 11.7 mph+ . In order to determine the direction, we use the Law of Sines:

1

sin sin 9010 11.6619

10sin 90sin 59.03611.6619

=

=

D

D D

So the direction is approximately 31D west of due north.


Observe that

Plane = 300cos30 , 300sin 30D D

Wind = 40cos120 ,40sin120D D Thus, the

Resultant = Plane + Wind 239.81, 184.64

Hence, Airspeed = |Resultant| 303 mph Heading = 1 184.6490 tan 52.41

239.81

D D

(East of due North) 65. The force required to hold the box in place is 500sin 30 250 lbs.=D 67. vertical component of velocity = 80sin 40 51.4 ft./sec.=D horizontal component of velocity = 80cos 40 61.3 ft./sec.=D 69. We need the component form for each of the three vectors involved. Indeed, they are:

0, 4 , 12,0 , 20cos330 , 20sin 330 17.32, 10A B C= = = D DG GG As such, 29.32, 6A B C+ + = G GG . So, ( )2229.32 6 29.93 yardsA B C+ + = + G GG .

60D

30D

Chapter 8

578


The given information yields a SAS triangle. So, we use the Law of Cosines:

2 2 2

2 2

2 cos400 100 2(400)(100)cos120170,000 80,000cos120

c a b ab = + = + =

DD

So, 458.26 lbs.c Next, we use the Law of Sines to find :

1

sin sin120100 458.26

100sin120sin 10.9458.26

=

=

D

D D


The given information yields a SAS triangle. So, we use the Law of Cosines:

2 2 2

2 2

2 cos1000 500 2(1000)(500)cos951,250,000 1,000,000cos951156.3545 1156 lbs.

c a b ab = + = + =

DD

75. The two forces acting on the hook are given in vector form by 200cos 45 ,200sin 45 100 2,100 2u = =D DG and 180,0v =G .

The resultant force is thus 100 2 180,100 2u v+ = +G G . The magnitude of this force is

( ) ( )2 2100 2 180 100 2 351.16u v+ = + + G G and the direction is

1100 2 100 2tan tan 23.75100 2 180 100 2 180

= = + + D .

60D

120D

Section 8.4

579


Using the information provided in the problem, we have

,0

, 5 cos 45 , sin 45

u u

R u R R

== = D DG GG G GG

As such, we have cos 45

sin 45 5

R u

R

= =

D

D

G GG

From the second equation, we see that 5

sin 455 2R = =DG

Using this in the first equation yields ( )5 2 cos 45 5u = =DG


Note that the resultant vector is the zero vector since the person starts and ends at the same point. To determine how far he or she walked, we need to find the lengths of all of the vectors pictured to the left, and sum them:

4,0 , so that 4

4,3 , so that 5

3, 3 , so that 3 2

3, 1 , so that 10

0, 1 , so that 1

OA OA

OA OA

OA OA

OA OA

OA OA

= == == == == =

JJJG JJJGJJJG JJJGJJJG JJJGJJJG JJJGJJJG JJJG

So, the total distance walked is ( )10 10 3 2 units+ + . 81. The torque is given by ( )45 0.2 sin85 8.97Nm = D . 83. The torque is given by ( )0.85 40 sin110 31.95Nm = D .

45D

Chapter 8

580


Observe that

1 2

cos150 , sin150 4 3, 4

cos 45 , sin 45 3 2,3 2

,

u u u

v v v

w w w

= = = ==

D D

D D

G G GG G GG

We want 0u v w+ + = GG G G . As such, we must have 4 3 3 2, 4 3 2w u v= = G G G . Note that

( ) ( )2

1

4 3 3 2 2 4 3 2

8.67

4 3 2 4 3 2tan tan4 3 3 2 4 3 3 2

71.95

w

= +

= = D

G

So, the direction of wG is 18.05D counterclockwise of SOUTH.

87. Magnitude cannot be negative. Observe that

( ) ( )2 22, 8 2 8 68 2 17 = + = = . 89. False. 1,0 1i = =G . 91. True. Let 1 2,u u u=G . Observe that ( ) ( ) ( )2 2 21 2 1 2 1 1 1,u u u u u u u u= = + = =G . 93. Vector (since need two pieces of information to precisely define this quantity).

95. ( )2 2 2 2,a b a b a b = + = + 97. Assume that u cv=G G , where c > 0, and that 1u =G . Then, we have

1c v =G , so that 1vc = G . Thus, ( )1vu v= GG G , as needed. 99. 126, 4 8, 4 2 1, 1 =

45D60D

Section 8.4

581

101. Observe that ( ) ( )

( )1 2 1 2 1 2

1 2 1 2 1 2

1 2 1 1 2 2

1 2 1 1 2 2

1 1 2 2

3 2 , 3 2 , ,

, 3 2 , 2 ,

, 3 2 , 2

, 6 3 ,6 3

6 2 ,6 26 2

u v u u u v v u u

u u v v u u

u u v u v u

u u v u v u

v u v uv u

+ = + = + = + = + = =

G G G

G G

103. Enter the following to compute the given quantity: [ [8] [-5] ] + 3[ [-7] [11] ] The output is: [ [-13] [28] ]. 105. First, compute the length of the vector, as follows: (102 + (-24)2) (output 26) Then, to obtain the unit vector, divide both components of the given vector by this length (26) to list as a FRACTION, enter the following:

1/26 [ [10] [-24] ] Frac Output is: [ [5/13] [-12/13] ] 107. Magnitude = Sum (( 33) ^ 2, (180) ^ 2) Direction angle = 1tan 180 ( 33) The output will be: Magnitude = 183 Direction Angle = -79.61114

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Young CAT2e SSM Ch8 Part1