chapter 6 triangle trigonometry - wikispacesman+ch+06.pdf · problem set 6-1 1. all measurements...

Problem Set 6-11. All measurements seem correct.

2. Answers may vary slightly.

Angle A Side a

0− 1.0 cm

30− 2.1 cm

60− 3.6 cm

90− 5.0 cm

120− 6.1 cm

150− 6.8 cm

180− 7.0 cm

3.

4.

No, the data don’t follow such a simple sinusoid.

5. The formula isthat is,

6. Answers will vary.

Problem Set 6-2

Q1. Q2.

Q3. 1 Q4.

Q5. Q6.

Q7. Sinusoidal axis

Q8.

Q9. Horizontal dilation by a factor of

Q10.

1.

2.

3.

4.

5. U = cosD1 32 + 42 D 22

2•3•4M 28.96−

k = √82 + 62 D 2•8•6 cos 172− M 13.97 m

r = √32 + 22 D 2•3•2 cos 138− M 4.68 ft

d = √72 + 92 D 2•7•9 cos 34− M 5.05 in.

r = √42 + 52 D 2•4•5 cos 51− M 3.98 cm

2 sin x cos x

15

+ sin 53− sin 42−cos 53− cos 42−

tanD1q

i√u2 D q2

u sin i

i

q

i

u

a2 = b2 + c2 D 2bc cos A.a2 = 32 + 42 D 2•3•4 cos A,

1234567

a (cm)

A

30° 90° 150°

y = 4 D 3 cos A

1234567

a (cm)

A

30° 90° 150°

6.

7.

8.

9. This is not a possible triangle, because .

10. This is not a possible triangle, because .

11. . Note that

so this is indeed a right triangle.

12. Note that

so this is indeed a right triangle.

13. a.

b.

14. a.

b. $2035.22

c. $2747.55

15.


17. so

18. a. If then whichhappens exactly when hence X is acute.If then whichhappens exactly when hence X is right.If then whichhappens exactly when so X is obtuse.

b. hence X is obtuse.

Problem Set 6-3Q1.

Q2.

Q3.

Q4.

Q5. The other values are either negative or greater than 180−,and therefore could not be angles of a triangle.

Q6. Amplitude

Q7. Start with the more complicated side and try to simplify itto equal the other side.

Q8. Cosine and secant

cos T =r2 + s2 D t2

2rs

t2 = r2 + s2 D 2rs cos T

s2 = r2 + t2 D 2rt cos S

r2 = s2 + t2 D 2st cos R

72 = 49 > 41 = 52 + 42;

cos X < 0;y2 + z2 D 2yz cos X > y2 + z2,x2 > y2 + z2,

cos X = 0;y2 + z2 D 2yz cos X = y2 + z2,x2 = y2 + z2,

cos X > 0;y2 + z2 D 2yz cos X < y2 + z2,x2 < y2 + z2,

= 42 + 52 D 2•4•5 cos Z= 42(sin2

Z + cos2 Z) + 52 D 2•4•5 cos Z

= 42 cos2

Z D 2•4•5 cos Z + 25 + 42 sin2

Zz2 = (4 cos Z D 5)2 + (4 sin Z D 0)2

Y = (5, 0),X = (4 cos Z, 4 sin Z),

cosD1 152 + 212 D 332

2•15•21M 132.2−

M 542.7249… ft150 + 200 + √1502 + 2002 D 2•150•200 cos 65−

G M 93−g = 8.0 cm,e = 6.0 cm,m = 5.0 cm,

R = 51−m = 5.0 cm,p = 4.0 cm,r M 4.0 cm,

15042 + 19532 = 24652,

Q = cosD1 15042 + 19532 D 24652

2•1504•1953M 90−.

14752 + 14282 = 20532,

O = cosD1 14752 + 14282 D 20532

2•1475•1428= 90−

6 + 3 < 12

7 + 5 < 13

E = cosD1 122 + 162 D 222

2•12•16M 102.64−

T = cosD1 62 + 72 D 122

2•6•7M 134.62−

G = cosD1 52 + 62 D 82

2•5•6M 92.87−

Precalculus with Trigonometry: Solutions Manual Problem Set 6-3 71© 2007 Key Curriculum Press

Chapter 6 Triangle Trigonometry

Q9.

Q10.

1.

2.

3.

4.

5.

6.

7.

8. a.

This is the same answer as in Problem 1.

b.

This is the same answer as in Problem 7.

9. a. so the triangle inequality shows that notriangle can have these three sides.

b.

According to Hero’s formula, the triangle would have tohave an impossible area. So no such triangle exists.

10. a.

b.

c.

11. a.

b.

A

0− 0.0000

15− 1.5529

30− 3.0000

45− 4.2426

60− 5.1962

75− 5.7956

90− 6.0000

105− 5.7956

120− 5.1962

135− 4.2426

150− 3.0000

165− 1.5529

180− 0.0000

c. False. The function increases from 0− to 90−, thendecreases from 90− to 180−.

θ

A = 12•4•3 sin θ = 6 sin θ

(0.06)(10,923) M $655

13,595

43,560 (35,000) M $10,923

12•150•200 sin 65− M 13,595 ft2

Area = √12(12 D 5)(12 D 6)(12 D 13) = √D504s = 1

2(5 + 6 + 13) = 12 cm

5 + 6 < 13,

Area = 12(2.4)(4.1)sin 63.1780…− = 4.3906… in.2

D = cosD12.42 + 4.12 D 3.72

2•2.4•4.1= 63.1780…−

= 5.4432… ft2

Area = √9.1606…(4.1606…)(0.1606…)(4.8393…)s = 1

2(5 + 9 + 4.3212…) = 9.1606… ftc = √52 + 92 D 2•5•9 cos 14− = 4.3212… ft

= √19.278 = 4.3906… in.2Area = √5.1(5.1 D 3.7)(5.1 D 2.4)(5.1 D 4.1)s = 1

2(3.7 + 2.4 + 4.1) = 5.1 in.

= √5,040,000 = 2244.9944… yd2

Area = √120(120 D 50)(120 D 90)(120 D 100)s = 1

2(6 + 9 + 11) = 13 cm

= √728 = 26.9814… cm2

Area = √13(13 D 6)(13 D 9)(13 D 11)s = 1

2(6 + 9 + 11) = 13 cm

12(34.19)(28.65)sin 138− M 327.72 yd2

12(4.8)(3.7)sin 43− M 6.06 cm2

12•8•4 sin 67− M 14.73 m2

12•5•9 sin 14− M 5.44 ft2

cos x cos y D sin x sin y

cos θ d. The figure is only a triangle with positive area for, so that is the domain. (We can sayif we consider the figure for or

to be a “flattened” triangle with area 0.)

12. a.

b.

c.

and there is no angle

that has a sine of 1.2987.

13. In , with , , , and ,we have and . Area formula:

.

(The formulas and give the same result.) Hero’s formula:

, so

14.

Problem Set 6-4Q1.

Q2.

Q3. 30−, 150− Q4. D0.372…

Q5. Q6. Scalene

Q7. Oblique Q8.

Q9. Q10. 4

1.

c =8 sin 97−sin 52− M 10.08 cm

b =8 sin 31−sin 52− M 5.23 cm

C = 180− D (52− + 31−) = 97−

cos2 x D sin2 x

sin2 θ + cos2

θ = 1

√3

2

1

2af sin P

p2 = a2 + f 2 D 2af cos P

A = 12bh = 1

2(5)(4) sin Zh = 4sin Z

h

4= sin Z

= √3

4=

√3

2

= √9 D 3

4•

3 D 1

4

= √3 + √3

2•3 D √3

2•

√3 + 1

2•

√3 D 1

2

= √3 + √3

2•

√3 + 1

2•3 D √3

2•

√3 D 1

2

= √3 + √3

2

(3 + √3

2D 1

)(3 + √3

2D √3

)(3 + √3

2D 2

) A = √s(s D a)(s D b)(s D c)

=3 + √3

2s =

1 + √3 + 2

2

12ab sin C1

2ac sin B

=√3

2=

1

2•√3•2•

1

2A =

1

2bc sin A

b = √3a = 1c = 2C = 90−B = 60−A = 30−∆ABC

sin θ =100

77M 1.2987,

77 sin θ = 100 cm2

θ = sinD1 77

77= 90−

77 sin θ = 77 cm2

θ = sinD1 50

77M 40.49− or 139.51−

1

2•14•11sin θ = 77sin θ = 50 cm2

θ = 180−θ = 0−0− ≤ θ ≤ 180−

0− < θ < 180−

72 Problem Set 6-4 Precalculus with Trigonometry: Solutions Manual© 2007 Key Curriculum Press

2.

3.

4.

5.

6.

7.

8.

9. a.

b. and

for a difference of $67,220.70.

c.for a savings of

$105,421.05 over y and $38,200.35 over x.

10. a. The internal angle at the turning point isso the angle at the start is

Then

ft.

b.

c. while

so it is faster to retrace the original route.

1125 ft

3 ft/s

= 375 s,399 ft

3 ft/s

+800 ft

5 ft/s

= 293 s

800 sin 137−sin 29− M 1125 ft

800 sin 14−sin 29− M 399

180− D (29− + 137−) = 14−.180− D 43− = 137−,

370d = $213,197.27,d = y sin 42− = 576.2088… m

370y = $318,618.32,370x = $251,397.62

y =1000 sin 58−

sin 80− = 861.1306… m

x =1000 sin 42−

sin 80− = 679.4530… m

Z = 180− D (42− + 58−) = 80−

w =500 sin 3−sin 175− M 300.24 m

l =500 sin 2−sin 175− M 200.21 m

O = 180− D (2− + 3−) = 175−

l =30 sin 87−

sin 8− M 215.26 ft

a =30 sin 85−

sin 8− M 214.74 ft

P = 180− D (85− + 87−) = 8−

w =5 sin 73−sin 59− M 5.58 ft

j =5 sin 48−sin 59− M 4.33 ft

A = 180− D (48− + 73−) = 59−

p =6 sin 28−sin 35− M 4.91 m

a =6 sin 117−

sin 35− M 9.32 m

F = 180− D (28− + 117−) = 35−

g =20 sin 99−

sin 2− M 566.02 km

i =20 sin 79−

sin 2− M 562.55 km

G = 180− D (2− + 79−) = 99−

s =120 sin 44−

sin 27− M 183.61 yd

h =120 sin 109−

sin 27− M 249.92 yd

S = 180− D (27− + 109−) = 44−

r =9 sin 34−sin 133− M 6.88 in.

p =9 sin 13−sin 133− M 2.77 in.

R = 180− D (13− + 133−) = 34− 11. a.

b.

c.

d. This is the complement of 51.3178…−

and one of the general values of

e. The principal values of go from 0− to 180−; anegative argument will give an obtuse angle and a positiveargument will give an acute angle, always the actual anglein the triangle. But the principal values of go fromD90− to 90−; a negative argument will never happen in atriangle problem, but a positive argument will only give anacute angle, while the actual angle in the triangle may bethe obtuse complement of the acute angle.

12.

The measured value should be

within 0.1 of 5.3 cm.


14.

So

and similarly.

Problem Set 6-5Q1. Side-Angle-Side

Q2.

Q3. The law of cosines

Q4.

Q5. Longest Q6. 180−

Q7. 5 Q8. A

Q9. 30− Q10. 150−

1.

or

2.

ft. (The other answer, D7.07 ft, is negative andtherefore impossible.)

3.

cm. (The other answer, D1.16 cm, is negative andtherefore impossible.)

4.

ft or 3.10 ftM 26.13

⇒ z =30 cos 13− ± √(D30 cos 13−)2 D 4•1•81

2•1

122 = z2 + 152 D 2•z•15 cos 13−

M 7.79

⇒ c =8 cos 34− ± √(D8 cos 34−)2 D 4•1•(D9)

2•1

52 = c2 + 42 D 2•c•4 cos 34−

M 16.82

⇒ z =10 cos 13− ± √(D10 cos 13−)2 D 4•1•(D119)

2•1

122 = z2 + 52 D 2•z•5 cos 13−

1.3169… cm= 5.3153… cm

⇒ c =8 cos 34− ± √(D8 cos 34−)2 D 4•1•7

2•1

32 = 42 + c2 D 2•4•c cos 34−

12•4•7 sin 38− M 8.62

√42 + 72 D 2•4•7 cos 38− M 4.57

x

sin X=

z

sin Z

x sin Z = z sin X

12 xy sin Z = 1

2 yz sin X

= 12 zx sin Y= 12

yz sin XA = 12 xy sin Z

x =(10.0) sin 30−

sin 110− M 5.3 cm.

180− D (40− + 30−) = 110−

arcsin x

arccos x

arcsin

10 sin A

7.

C = cosD1 42 + 72 D 102

2•4•7= 128.6821…−

C = sinD1

10 sin A

7= 51.3178…−

A = cosD1 42 + 102 D 72

2•4•10= 33.1229…−


5.But . The discriminant isnegative. There is no solution. Side b is too short.

6.

But The discriminantis negative. There is no solution. Side x is too short.

7.

in. (The other answer, D26.09 in., is negative andtherefore impossible.)

8.

m. (The other answer, D29.74 m, is negative andtherefore impossible.)

9. a. By alternate interior angles, the angle at Ocean City is also 50−, so

b. The other answer is mi. This means 12.94 milesto the west of Ocean City.

c. Let be the angle at the easternmost range along thebeach and let K be the angle at KROK. Rather than findingK directly using the approximate answer we just got, usethe exact given distances to first find

Then (which matchesthe angle on the map).

10. .

Because there is another solution,

11. First, find

Because there is only one solution.

12. .


13. .


14. a. b.

c. d.

e. f.

Problem Set 6-6Q1. B Q2. A

Q3. Q4. 10 ft2

Q5. 60 cm Q6. Harmonic analysis

d2 + e2 D 2de cos F

y sin X < y < xx < y sin X < y

y sin X < y < xy sin X < x < y

x = y sin X < yx < y sin X < y

b > g,

G = sinD1

900 sin 110−1000

M 57.75−

x > z,

Z = sinD1

(7.5)sin 58−9.3

M 43.15−

h > c,S M 180− D 28− D 10.82− = 141.18−

C = sinD1

20 sin 28−50

M 10.82−.

C M 180− D 23.00− = 157.00−.a < c,

C = sinD1

30 sin 19−25

M 23.00−

K M 180− D (50− + 30.71−) = 99.29−

M 30.71−.⇒ θ = sinD1

20 sin 50−30

sin θ

20=

sin 50−30

θ.

θ

M D12.94

M 38.65 mi.

⇒ x =40 cos 50− ± √(D40 cos 50−)2 D 4•1•(D500)

2•1

302 = x2 + 202 D 2•x•20 cos 50−

M 8.07

⇒ b =22 cos 170− ± √(D22 cos 170−)2 D 4•1•(D240)

2•1

192 = b2 + 112 D 2•b•11 cos 170−

M 5.52

⇒ s =32 cos 130− ± √(D32 cos 130−)2 D 4•1•(D144)

2•1

202 = s2 + 162 D 2•s•16 cos 130−

(D120 cos 13−)2 D 4•1•3456 M D152.68.⇒ z2 + (D120 cos 13−)z + 3456 = 0122 = z2 + 602 D 2•z•60 cos 13−

(D8 cos 34−)2 D 4•1•12 M D4.01c2 + (D8 cos 34−)c + 12 = 022 = c2 + 42 D 2•c•4 cos 34− ⇒ Q7. Q8.

Q9. Exponential

Q10. Horizontal dilation by a factor of

1.

2.

3.

4. a.mi/h

The resultant could equal only if the velocitieswere in the same direction.

b.

5. a.

Lucy’s bearing is .

b. The starting point’s bearing from Lucy is.

c.

6. a. km/h

b. Your speed must be km/h.

c. No. Any upstream component of your 3 km/h velocity can never cancel the 5 km/h downstream component of the water.

7.

8.

9.

10.

11. a.

b.units

= D76.7360…− + 180n− = 103.2640…−

θ = tanD154.3745…

D12.8175…+ 180n−

= 55.8648…|rA

| = √(D12.8175…)2 + (54.3745…)2

= D12.8175… iA

+ 54.3745… jA+ (21 sin 70− + 40 sin 120−)jA

(21 cos 70− + 40 cos 120−)iA797.4644… i

A+ 306.1179… j

AD6.1344… i

A+ 14.4519… j

AD1782.0130… i

A D 907.9809… jA

6.0376… iA D 5.2484… j

A

5 cot 34− = 7.4128…

θ = tanD1

5

3= 59.0362−

|rA

| = √32 + 52 = √34 = 5.8309…

√1002 + 1802 = √42,400 = 205.9126… m

360− D 29.0546…− = 330.9453…−

180− D 29.0546…− = 150.9453…−

θ = tanD1

100

180= 29.0546…−

M 11.10−

α = cosD14002 + (521.2268…)2 D 1502

2•400•(521.2268…)

400 + 150 = 521.2268… M 521.23

|rA

| = √4002 + 1502 D 2•400•150 cos 138−

M 150.00−

α = cosD1 92 + (11.6931…)2 D 202

2•9•(11.6931…)

= 11.6931… M 11.69 in.|aA

+ bA

| = √92 + 202 D 2•9•20 cos 17−θ = 180− D 163− = 17−

α = cosD1 82 + (9.5995…)2 D 22

2•8•(9.5995…)M 7.86−

= 9.5995… M 9.60 ft|aA

+ bA

| = √82 + 22 D 2•8•2 cos 139−θ = 180− D 41− = 139−

α = cosD1 72 + (14.6637…)2 D 112

2•7•(14.6637…)M 45.84−

= 14.6637… M 14.66 cm|aA

+ bA

| = √72 + 112 D 2•7•11 cos 107−θ = 180− D 73− = 107−

1

3

7

11

12

13


12. a.

b.units

13.

14.

mi

15.

mi/h

16.

ft/min

17.

18. a.

5−5

5

−5

a + b

b

a

aA

+ bA

= (5 D 4)iA

+ (2 + 3)jA

= iA

+ 5jA

θ = tanD1

51.2124…

6.8478…= 82.3838…−

= 51.6682… newtons|rA

| = √(6.8478…)2 + (51.2124…)2= 6.8478… i

A+ 51.2124… j

A+ (90 sin 40− + 50 sin 110− + 70 sin 230−)jArA

= (90 cos 40− + 50 cos 110− + 70 cos 230−)iA = D21.8416…− + 180n− = 158.1584…−

θ = tanD144.5540…

D111.1593…+ 180n−

= 119.7558…|rA

| = √(D111.1593…)2 + (44.5540…)2= D111.1593… i

A+ 44.5540… j

A+ (100 sin 170− + 30 sin 115−)jA(100 cos 170− + 30 cos 115−)iA

= D55.8929…− + 180n− = 304.1070…−

θ = tanD1D138.9784…

94.1204…+ 180n−

= 167.8484…|rA

| = √(94.1204…)2 + (D138.9764…)2= 94.1204… i

A D 138.9764… jA+ (200 sin 320− + 60 sin 190−)jA

(200 cos 320− + 60 cos 190−)iA = D16.4841…− + 180n− = 343.5158…−

θ = tanD1D3.3146…

11.2015…+ 180n−

= 11.6816… |rA

| = √(11.2015…)2 + (D3.3146…)2= 11.2015… i

A D 3.3146… jA+ (30 sin 200− + 40 sin 10−)iA

(30 cos 200− + 40 cos 10−)iA= 41.7867…− + 180n− = 41.7867…−

θ = tanD146.6452…

52.1940…+ 180n−

= 70 mi|rA

| = √(52.1940…)2 + (46.6452…)2= 52.1940… i

A+ 46.6452… j

A+ (50 sin 20− + 30 sin 80−)jA(50 cos 20− + 30 cos 80−)iA

= 20.9409…− + 180n− = 20.9409…−

θ = tanD14.2639…

11.1423…+ 180n−

= 11.9303…|rA

| = √(11.1423…)2 + (4.2639…)2= 11.1423… i

A+ 4.2639… j

A+ (12 sin 60− + 8 sin 310−)jA(12 cos 60− + 8 cos 310−)iA b.

c. The resultant vector is the same in both cases.

19.

20. The magnitude is 0; the direction is undefined. The resultantis the vector the zero vector.

21. If and are any two vectors, then a, b, c, andd are real numbers. So and are also real numbers,since the real numbers are closed under addition. Therefore,the sum exists and is a vector, so the set ofvectors is closed under addition. The zero vector is necessaryso that the sum of any vector and its opposite,

will exist.

22. If is any vector, then a and b are real numbers. So, if c is any scalar, i.e., a real number, then ca and cb are realnumbers. So the product exists and is a vector.Therefore, the set of vectors is closed under scalarmultiplication. The zero vector is necessary so that theproduct of any vector with the scalar 0 will exist.

23. Scalar is from the Latin meaning “ladder”.

Problem Set 6-7Q1.

Q2. Q3. 12 bc sin A

a

sin A=

C

sin C

b2 = a2 + c2 D 2ac cos B

scalae,

ca iA

+ cb jA

a iA

+ b jA

Da iA

+ b jA,

a iA

+ b jA

(a + c)iA

+ (b + d)jA

b + da + cc iA

+ d jA

a iA

+ b jA

0 iA

+ 0 jA,

5−5

5

−5

a + (b + c)

b + c

bc

a

5−5

5

−5

a + b

(a + b) + c

b

ca

5−5

5

−5

b

a

b + a

bA

+ aA

= (D4 + 5)iA

+ (3 + 2)jA

= iA

+ 5jA


Q4.

Q5. Q6.

Q7. Q8. C.

Q9.

Q10. D37−

1. Let A be the point from which the angle is 21.6−, B the pointfrom which the angle is 35.8−, C the top of the mountain, and D the foot of the altitude. By a theorem of geometry,

(an exterior angle of a triangle equalsthe sum of the opposite interior angles), so

Then, by the law of sines,

so

2. a.

Window

Roof

b. Area

3. Let x be the length of the first leg of the detour and y be thelength of the second leg.

a.

b.

4. a.

The shelter will be able to display about

.

b. θ = cosD1 402 + 1002 D 702

2•40•100= 33.1229…−

1092.9

3M 364 pumpkins

= √1194375 M 1092.9 ft2

Area = √105(105 D 40)(105 D 70)(105 D 100)s = 1

2(40 + 70 + 100) = 105 ft

A =1

2•70•x sin 21− M 607.5 km2

(x + y) D 70 M 8.7 km

y =70 sin 21−sin 124− M 30.3 km

x =70 sin 35−sin 124− M 48.4 km

180− D (21− + 35−) = 124−

M 120.6 ft2

=1

2•22•

(22 sin 65−

sin 82−

)•sin 33−

=22 sin 65−

sin 82− M 20.1 ft

=22 sin 33−

sin 82− M 12.1 ft

180− D (33− + 65−) = 82−

M 445.1 m

=507 sin 21.6− sin 35.8−

sin 14.2−CD = BC sin 35.8−

BC =507 sin 21.6−

sin 14.2−

= 35.8− D 21.6− = 14.2−.∠ACB = ∠CBD D ∠CAB

∠CBD = ∠CAB + ∠ACB

sin A cos B D cos A sin B

D1D2iA

+ 15jA

a

bv

a + b

a

b

vA

= aiA

+ bjA

X

Z

z Y

yx x

5.

or 608.4 yd

6. First find the angle at the peak, and then find the desired angle as But

which is not the sine of any angle. It is impossible to buildthe truss to the specifications. The 20-ft side is too short, orthe 30-ft side is too large, or the 50− angle is too large.

7. Let A be the observer, B the launching pad, C the missilewhen at 21−, and D the missile when at 35−.a.

b. km/s

c.

8.

or 35.2 in.

9. a.

km/h

b.

km/h

10. a. Lift:Horizontal component:

L (lb) H (lb)

0− 500,000 0

5− 501,910 43,744

10− 507,713 88,163

15− 517,638 133,975

20− 532,089 181,985

25− 551,689 233,154

30− 577,350 288,675

b. The centripetal force is stronger, so the plane is beingforced more strongly away from a straight line into acircle.

c. The horizontal component is 0, so there is no centripetalforce to push the plane out of a straight path.

d.

e. Most important, the plane would start to fall, because thevertical component would be less than 500,000 lb andcould not support it. Together with the turning caused bythe horizontal component, this would result in a spiraldownward.

θ = cosD1 500,000

600,000M 33.56−

θ

H = 500,000 tan θL = 500,000 sec θ

M 463.4|rA

| = √5002 + 402 D 2•500•40 cos 23−θ = 23−

M 537.0|rA

| = √5002 + 402 D 2•500•40 cos 157−θ = 180− D 23− = 157−

M 63.7 in.

x =110 cos 26− ± √(D110 cos 26−)2 D 4•1•2241

2•1

x2 + (D110 cos 26−)x + 2241 = 0282 = x2 + 552 D 2•x•55 cos 26−

tanD1

2.6657…

2M 53.1210…−

= 2.6657… km= 10(0.1265…) + 2 tan 35−10(0.1265…) + BD

0.6327… km

5 s

= 0.1265…

= 0.6327… km= 2 tan 35− D 2 tan 21−CD = BD D BC

sin θ =30 sin 50−

20M 1.15,

180− D (50− + θ).θ,

M 1380.6 yd

2000 cos 6− ± √(D2000 cos 6−)2 D 4 • 1 • 840,000

2 • 1x =

x2 + (D2000 cos 6−) x + 840,000 = 04002 = x2 + 10002 D 2•x•1000 cos 6−


11. Let F H the other person’s force. Since the verticalcomponents must cancel out, so

lb.

Then lb.

12.

13.

14. a.

b. The largest angle is at the space station.

c. The remaining angle of the triangle is

15. Let A be the center of Earth, B be where the line from you tothe center of Earth intersects the surface of Earth, C be you,and D be the horizon.

km

rad

km

16. a.

or

b.

but , so there is no possible solution. Or simply note that when the height of the hinge is

which is greater than the length of the second ruler.

c.

17. a.

b.

M 6838.2 m2

√268(268 D 114)(268 D 165)(268 D 257)

114 + 165 + 257

2= 268 m

cosD1 1142 + 1652 D 2572

2•114•165M 133.4−

100 sin θ = 60 cm ⇒ θ = sinD1 0.6 = 36.8698…−

100 sin 50− = 76.6044… cm,θ = 50−,

(D200 cos 50−)2 D 4•1•6400 = D9072.9633… < 0

x =200 cos 50− ± √(D200 cos 50−)2 D 4•1•6400

2•1

44.6719… cm = 143.2665… cm

x =200 cos 20− ± √(D200 cos 20−)2 D 4•1•6400

2•1

= 6400(0.0558…) M 357.5BD→

= rθ

= 3.2008π

180= 0.0558…

θ = tanD1357.9106…

6400= 3.2008…−

= 357.9106…CD = √64102 D 64002

AC = 6400 + 10 = 6410 kmAD = AB = 6400 km

b =sin 113−•4362

sin 29.3− M 8205 km

180− D 37.7− D 113− = 29.3−.

Area = √11(11 D 5)(11 D 7)(11 D 10) M 16.2 km2

s = 12(5 + 7 + 10) = 11 km

θ = cosD1 52 + 72 D 102

2•5•7M 111.8−

√82 + 112 D 2•8•11 cos 120− = √273 M 16.5 km

θ M tanD1223.1

186.4M 50.1−

|rA

| M √186.42 + 223.12 M 290.7 km/hM 186.4i

A+ 223.1j

A+ (52 sin 15− D 250 sin 237−)jArA

= (52 cos 15− D 250 cos 237−)iA

M D13.6− + 180n− = 166.4−θ M tanD1 5.9

D24.4

M 25.1 knots|rA

| M √(D24.4)2 + 5.92

M D24.4iA

+ 5.9jA+ (22 sin 157− + 5 sin 213−)jA

rA

= (22 cos 157− + 5 cos 213−) iAM 110.850 cos 20− + (66.0732…) cos 15−

= 66.0732… M 66F =50 sin 20−

sin 15−

F sin 15− = 50 sin 20−,18. Let the 50-m side be AB, the 60-m side be BC,

the 70-m side be CD, and the remaining side be DA.

a.

m

so

b.

c. so

19. a. Answers will vary.

b. The program should give the expected answer.

c. Label the 95− angle A, and label the rest of the verticesclockwise as B through F.

m

m

m

m

d. For a nonconvex polygon, you might not be able to divideit into triangles that fan out radially from a single vertex.

Problem Set 6-8

Review Problems

R0. Journal entries will vary.

R1. a. Answers may vary slightly.

Third Side (cm)

30− 2.5

60− 4.6

90− 6.4

120− 7.8

150− 8.7

b. 5 + 4 = 9; 5 D 4 = 1

θ

= 30.6817…AF = √AE 2 + 172 D 2•AE•17 cos ∠AEF∠AEF = 115− D ∠AED = 34.5489…−

∠AED = sinD1

AD sin ∠ADE

AE= 80.4510…−

= 43.1295…AE = √AD2 + 182 D 2•AD•18 cos ∠ADE∠ADE = 122− D ∠ADC = 75.6949…−

∠ADC = sinD1

AC sin ∠ACD

AD= 46.3050…−

= 43.8929…AD = √AC2 + 152 D 2•AC•15 cos ∠ACD∠ACD = 147− D ∠ACB = 115.7712…−∠ACB = sinD1

20 sin 114−AC

= 31.2287…−

= 35.2410…AC = √202 + 222 D 2•20•22 cos 114−

M 58.0−∠BAD = 360 D (127− + 132− + ∠ADC)

∠ADC = sinD1

AC sin ∠ACD

AD 43.0−

sin ∠ADC

AC=

sin ∠ACD

AD,

= 137.5 mAD = √702 + AC2 D 2•70•AC cos ∠ACD

M 4476.4 m2AABCD = A∆ABC + A∆ACD

= 3278.4517… m2

A∆ACD =1

2AC•70 sin ∠ACD

∠ACD = 132− D ∠ACB = 108.0951…−

∠ACB = sinD150 sin 127−

AC= 23.9048…−

sin ∠ACB

50=

sin 127−AC

,

= 98.5438…AC = √502 + 602 D 2•50•60 cos 127−

= 1197.9532… m2

A∆ABC =1

2•50•60 sin 127−


c. ; yes

d.

No, the shape is not a sinusoid. (The lower curve, whichthe students do not have to include, is the sinusoid withthe same starting and ending points, .)

R2. a.

b.

c. Also, whichever angle we try to calculate, weget an impossible cosine:

d.

= e2 + f 2 D 2ef cos θ = e2 (cos2

θ + sin2 θ) + f 2 D 2 ef cos θ

= e2 cos2

θ D 2ef cos θ + f 2 + e2 sin2θd2 = (e cos θ D f )2 + (e sin θ D 0)2F = (e cos θ, e sin θ)E = (f, 0)

D (0, 0) E (f, 0)f

e d

F

u

v

102 + 32 D 52

2•10•3= 1.4

52 + 102 D 32

2•5•10= 1.16

32 + 52 D 102

2•3•5= D2.2

3 + 5 < 10.

cosD1 82 + 52 D 112

2•8•5M 113.6−

113.6°

811

5

M 77.9 ft√502 + 302 D 2•50•30 cos 153−

50

30

153°77.9

5 D 4 cos θ

2

4

6

8

Third side (cm)

θ

60° 120° 180°

√52 + 42 M 6.4 R3. a.

b.

mi2

mi2

c.

or 138.2−

d.

R4. a.

b.

5 sin 112−sin 30− M 9.3 m

5

112°38°

6 sin 48−sin 39− M 7.1 in.

7.1

6

48°

39°

= 12de sin F

A = 12bh

base = d, altitude = e sin F

D

EF

f

d

e

θ = sinD12•40

10•12M 41.8−

12•10•12 sin θ = 40

= 42.84857057…A = √17(17 D 8)(17 D 11)(17 D 15)

s =8 + 11 + 15

2= 17

A = 12•8•11 sin θ = 42.84857057…

= 103.1365587…−

θ = cosD1 82 + 112 D 152

2•8•11

11

15

8103.13°

5030

153°77.9

12•50•30 sin 153− M 340.5 ft2


c.

or 133.9−

d.

and similarly.

R5. a.

cm or 3.4 cm

b. , which is not the sine of any angle.

c. The 5-cm side must be perpendicular to the third side,making the 8-cm side the hypotenuse of a right triangle.

Then

d. As in part a, but with

cm. (The other answer, cm, wouldrepresent the triangle with the 5-cm side to the left of the8-cm side.)

R6. a.

(This is since by

inspection the angle is obtuse.)

b.

since (12, D3) is in the fourth quadrant.

c.

d.

km/h

θ = tanD1 D138.4578…

D255.1704…M 28.5− + 180n− = 208.5−

M 290.3

√(D255.1704…)2 + (D138.4578…)2|rA

| == (D255.1704…)i

A+ (D138.4578…)j

A+ (300 sin 220− + 60 sin 115−)jA(300 cos 220− + 60 cos 115−)iAθ = tanD1

33.2088…

D128.5575…M 165.5−

M 132.8 mi|rA

| = √(D128.5575…)2 + (33.2088…)2+ (33.2088…)j

A= (D128.5575…)i

A+ (120 sin 270− + 200 sin 130−)jA= (120 cos 270− + 200 cos 130−)iAa

A+ b

A

θ = tanD1 D312 M 346.0−,

|rA

| = √122 + (D3)2 = √153 M 12.4aA

+ bA

= (5 + 7)iA

+ (3 D 6)jA

= 12iA D 3j

A

180− D 14.8−,φ = sinD110 sin 6−

|rA|M 165.2−.

|rA

| = √62 + 102 D 2•6•10 cos 6− = 4.0813… θ = 180− D 174− = 6−

M D3.7M 10.5

x =10 cos 47− ± √(D10 cos 47−)2 + 4•1•39

2•1

φ = 47−

M 38.7−.θ = sinD1 5

8

sin φ =8 sin 85−

5M 1.6

M 11.4

x =16 cos 22− ± √(D16 cos 22−)2 D 4•1•39

2•1

x2 + (D16 cos 22−)x + 39 = 052 = x2 + 82 D 2•x•8 cos 22−

e

sin E=

f

sin F

e sin F = f sin E

12de sin F = 1

2df sin E

sinD1

7 sin 31−5

M 46.1−

31°7

55

R7. a. so it isout of range.

b.

km or km

c., so x is undefined.

d. The line from the plane to Tokyo Airport must beperpendicular to the flight path, so

km.

e.The theorem states that if P is a point exterior to circle C,PR cuts C at Q and R, and PS is tangent to C at S,then

f.

from Nagoya Airport

from Tokyo Airport

Nagoya Airport is closer by .

g.

h.

i. The helicopter can tilt so that the thrust vector exactlycancels the wind vector.

Concept Problems

C1. Student essay

C2. a.

The answers are the same because.

b.

The answers are opposite because

c.

Directly: First find

C3. Student project

1

2•10•12 sin C +

1

2•6•7 sin 250− M 30.7 ft2

C = cosD1

159 + 84 cos 250−240

= 57.1260…−

240 cos C = 159 + 84 cos 250−= 102 + 122 D 2•10•12 cos C

DB2 = 62 + 72 D 2•6•7 cos 250−∠C.

AABCD = A∆BCD D A∆ABD M 30.7 ft2

= 50.3919… ft2

A∆BCD = √s(s D 10)(s D 12)(s D DB)

s =10 + 12 + DB

2= 16.3322… ft

sin 250− = Dsin 110−.12•6•7 sin 250− = D19.7335… ft2

12•6•7 sin 110− = 19.7335… ft2

cos 250− = cos 110−

√62 + 72 D 2•6•7 cos 250− M 10.7 ft

√62 + 72 D 2•6•7 cos 110− M 10.7 ft360− D 250− = 110−

√30002 + 4002 M 3026.5 lb

tanD1 400

3000M 7.6−

M 35.2 km

260 sin 35−sin 118− M 168.9 km

260 sin 27−sin 118− M 133.7 km

180− D (35− + 27−) = 118−PQ•PR = PS2.

2402 = 57,600 = (177.17…)(325.111…)

θ = sinD1100

260M 22.6−x = √2602 D 1002 = 240

= D71,722.76…(D520 cos 40−)2 D 4•1•57,600

325.111…=177.1700…

520 cos 15− ± √(D520 cos 15−)2 D 4 • 1 • 57,600

2 • 1x =

x2 + (D520 cos 15−)x + 57,600 = 01002 = x2 + 2602 D 2•x•260 cos 15−

√2602 + 2202 D 2•260•220•cos 32− M 137.8 km,


C4. a. Sketch should match Figure 6-8h.

b. Because each is a radius of the circle, let By the Pythagorean property,(1) By the law of cosines,(2) (3) Substituting (1) into (2) and (3) and rearranging, (4) (5) Dividing (5) into (4),

By cross multiplication, rearranging, and factoring,

Chapter Test

T1.

T2. or

T3.

T4. ASA is shown, but the law of cosines works only for SAS and SSA.

T5. SAS is shown, but the law of sines works only for ASA, SAA,and SSA.

T6. Also, if we try to use the law of cosines to findany of the angles, we get

or

none of which is the cosine of any angle.

T7. The range of is which includes everypossible angle measure for a triangle. But the range of is so the function cannot find obtuseangles.

T8.

T9.

−5

3

x

y

−5j 3i – 5j

3i

a + ba

b

sinD1D90− ≤ θ ≤ 90−,sinD1

0− ≤ θ ≤ 180−,cosD1

192 + 72 D 102

2•19•7M 1.2,

102 + 192 D 72

2•10•19M 1.1,

72 + 102 D 192

2•7•10M D1.5,

10 + 7 < 19.

A = 12de sin C

sin C

c=

sin D

d=

sin E

e

c

sin C=

d

sin D=

e

sin E

d2 = c2 + e2 D 2ce cos D

PQ•PR = PS2

PQ•PR (PR D PQ) = PS2(PR D PQ)PQ•PR2 D PR•PQ2 = PR•PS2 D PQ•PS2

PQ•PR2 + PQ•PS2 = PR•PQ2 + PR•PS2

PQ

PR=

PQ2 + PS2

PR2 + PS2

2(PR)(PO) cos α = PR2 + PS2

2(PQ)(PO) cos α = PQ2 + PS2

r2 = PR2 + PO2 D 2(PR)(PO) cos αr2 = PQ2 + PO2 D 2(PQ)(PO) cos α

PO2 = PS2 + r2

SO, QO, RO = r.

T10. Student drawing. The third side should be about

T11.

T12.

T13.

T14, T15, T16. Answers will vary.

T17.

cm or 2.445612733… cm

T18.

T19.

T20.

since (3, D5) is in the fourth quadrant.

T21. Student essay

Problem Set 6-9

Cumulative Review, Chapters 1–6

1. where

2. Horizontal dilation by , vertical dilation by 5

3. Horizontal translation by 3, vertical translation by

4.

5. Odd

6. g(x) = 3f c12

(x D 4)d + 5

1

1

y

x

f −1(x)

f (x)

h(x) = f (x D 3) D 2D2;

13

a ≠ 0f (x) = ax2 + bx + c

θ = tanD1 D5

3M 301.0−,

|rA

| = √32 + (D5)2 = √34 M 5.8

(6.54232772…)(2.445612733…) = 16 = 42

√52 D 32 = 4 cm

= 6.54232772…

x =10 cos 26− ± √(D10 cos 26−)2 D 4•1•16

2•1

x2 + (D10 cos 26−)x + 16 = 032 = x2 + 52 D 2•x•5 cos 26−

50 sin 38−sin 95− M 30.9 ft

180− D (38− + 47−) = 95−50

38° 47°

√72 + 52 D 2•7•5 cos 24− M 3.2 cm

3.2 cm.


7.

Assume that the observer is on the equator, the satellite is inan equatorial circular orbit, and the rotational speeds ofEarth and the satellite are each constant. This is the graph of

where r is the radius ofEarth and h is the height of the satellite.

8.

9.

10. Reference angle third quadrant;

11.

the u-coordinate on the x-axis 12.

13. Sinusoidal

14.

15. 2•180−

π= 114.5915…−

2π; π; π

2; π

4

1

−1

y = sin θ

θ

90 270°

= D1cos 180− =

(−1, 0)180°

u

v

sin 240− = D√3

2

= 240− D 180− = 60−,

csc θ = D 135sec θ = 13

12,cot θ = D 125 ,

tan θ = D 512,cos θ = 12

13,sin θ = D 513,

√122 + (D5)2 = 13;

−213°

33° u

v

d = √r2 + (r + h)2 D 2r (r + h) cos t,

t

d

16.

17.

18. a. b. 4

c. D6 d. 3

19. a. times horizontal dilation is the period.

b. Amplitude

c. Phase displacement or phase shift

d. Sinusoidal axis

20. Sinusoidal axis

AmplitudePeriodHorizontal dilationPhase shift H 1Starts high (cosine)

21. H 3.4452…

22.

23.

510

1520

2

4

y

x

M 2.3, 9.7, 12.3, …x = 1 ± 5

π cosD1 23 + 10nx D 1 = ± 5

π cosD1 23 + 10n

π5 (x D 1) = ± cosD1 23 + 2nπcos

π5 (x D 1) = 2

3

3 cos π5 (x D 1) = 2

2 + 3 cos π5 (x D 1) = 4

y = 2 + 3 cos π5 (342.7 D 1)

y = 2 + 3 cos π5 (x D 1)

(B = π5)= 10

2π = 5π

= 11 D 1 = 10= 5 D 2 = 3

=5 D 1

2= 2

2π

1

5

1

−1

y = cos x

θ

180° 360°

2

1

−1

v

u

x

2 rad

−1

−1

1

2


24. Sinusoidal axis

AmplitudePeriodHorizontal dilationPhase shiftThe function starts low, so you use the negative cosinefunction.

25. a. ;

H 229.1831… rev/min

b. 120 cm/s

c. H

95.4929… rev/min

26. a. H 0.0628… rad/min

b.

c. The angular velocity is the coefficient, B, of the argument.

27. Reciprocal properties:

Quotient properties:

Pythagorean properties:

28.

For ( )

29.Cosine of first, cosine of second, plus sine of first, sine of second

30.

31.

32.

since

is first quadrant

33.

34. so which is a sinusoid.

35. Larger sinusoid:

Amplitude: Period: 60−

Horizontal dilation: (B H 6)

Phase shift: 0Starts high: cosiney = 3 cos 6θ

60−360− =

1

6

3 D 0 = 3 m

sin2x = 12 D 1

2cos 2x,cos 2x = 1 D 2 sin2x,

= 12 sin θ cos θ= 6•2 sin θ cos θ= 6(sin θ cos θ + cos θ sin θ)

6 sin 2θ = 6 sin(θ + θ)

N 3 cos θ + 4 sin θ = 5 cos(θ D 53.1301…−)

(3

5, 4

5

)D = cosD1

3

5= sinD1

4

5= 53.1301…−,

A = √32 + 42 = 5

cos(34−) = sin(90− D 34−) = sin 56−= 0•cos θ + 1•sin θ = sin θ;cos(90− D θ) = cos 90− cos θ + sin 90− sin θ

cos 34− = 0.8290… = sin 56−

cos(x D y) = cos x cos y + sin x sin y

x ≠ 90− + n•180−cos x ≠ 0

= sin2

x

cos4 x

= sin2

x(cos2 x + sin2

x)

cos4 x

= sin2

x cos2 x + sin4

x

cos4 x

=sin2

x cos2 x

cos4 x

+sin4

x

cos4 x

sec2 x sin2

x + tan4 x =

sin2 x

cos2 x

+sin4

x

cos4 x

1 + cot2 θ = csc2

θtan2 θ + 1 = sec2

θ,sin2 θ + cos2

θ = 1,

cot θ =cos θ

sin θtan θ =

sin θ

cos θ,

cot θ =1

tan θcsc θ =

1

sin θ,sec θ =

1

cos θ,

M 18,850 mi/h π rad

50 min•

5000 mi

rad•

60 min

h= 6000π mi/h

1 rev

100 min•

2π rad

rev=

π

50 rad/min

120 cm

s•

1 rev

2π•12 cm•

60 s

min=

300

π rev/min

24 rad

s•

1 rev

2π rad•

60 s

min=

720

π rev/min

120 cm

s•

1 rad

5 cm= 24 rad/s

d = 5000 D 4000 cos π50t

= 0(B = π

50)= 1002π = 50

π

= 2•50 = 100= 9000 D 5000 = 4000

=1000 + 9000

2= 5000 Smaller sinusoid:

Amplitude:

Period:

Horizontal dilation: (B H 30)

Phase shift: 0Starts at 0: sine

Combined:

Varying sinusoidal axis: add

36. Larger sinusoid:

AmplitudePeriod

Horizontal dilation (B H 1)

Phase shiftStarts at 0: sine

Smaller sinusoid:

Amplitude

Period

Horizontal dilation

Phase shift H 0Starts high: cosine

Combined:

Varying amplitude: multiply

37.

38. In the problem, the period of is 18−; the period ofis 360−. These are much different. In the answer, the

period of is 17.1428…−; the period of is18.9473…−. These are nearly equal.

39.

40. Principal value: radComplement: radGeneral: or

41. See Figure 6-9h in the student text.Possible parametric equations:

42. Domain is Range is 0

43.243.4−, 423.4−, 603.4−

44. In (and similarly for and ). The square of one side of a triangle is the sum of thesquares of the other two sides minus twice their producttimes the cosine of the angle between them.

45. In The length of one side of a

triangle is to the sine of the angle opposite it as the length ofany other side is to the sine of the angle opposite that side.

46.The area of a triangle is the product of any two sides andthe sine of the angle between them.

47. cosD162 + 72 D 122

2•6•7M 134.6−

12

= 12 ca sin B.= 1

2 bc sin AA∆ABC = 1

2 ab sin C

a

sin A=

b

sin B=

c

sin C.∆ABC,

b2a2c2 = a2 + b2 D 2ab cos C∆ABC,

N θ M 63.4−,θ = arctan 2 = 63.4349…− + 180n−

− ≤ y ≤ 180−D1 ≤ x ≤ 1

x = cos t, y = t

2.7300… + 2nπ rady = 0.4115… + 2nπ radπ D 0.4115… = 2.7300…

y = sinD1 0.4 = 0.4115…

θ = 78.6900…−

cos 19θcos 21θcos θ

cos 20θ

= cos 21θ + cos 19θy = cos(20θ + θ) + cos(20θ D θ)

y = 5 sin θ cos 12θ

y = cos 12θ

=360−/12

360− =1

12 (B = 12)

=360−12

= 1

y = 5 sin θ

= 0

=360−360− = 1

= 360−= 5

y = 3 cos 6θ + 2 sin 30θ

y = 2 sin 30θ

12−360− =

1

30

60−5

= 12−

5 D 3 = 2


48.

49. a.

b.

since (2, 16) is in the first quadrant.

c.

d. False. This is true only if and are at the same angle.bA

aA

2−2−4

5

10

15

u

v

a + bb

a

aA

, bA

, aA

+ bA

θ = tanD116

2M 82.9−,

|rA

| = √22 + 162 = √260 M 16.1

(D3 + 5)iA

+ (4 + 12)jA

= 2iA

+ 16jA

= 14.9478… ft2

Area = √12.5(12.5 D 6)(12.5 D 7)(12.5 D 12)s = 1

2(6 + 7 + 12) = 12.5 ft 50. In units of 1000 miles:

a.

b. As in Problem 24 and part a, so .

c. The dashed curve represents the equation from Problem 24.

51. Student essay

50 100

5

t (min)

y (1000 mi)

y = √41 D 40 cos π50t

B = π50,

= √41 D 40 cos xy = √42 + 52 D 2•4•5 cos x


chapter 6 triangle trigonometry - wikispacesman+ch+06.pdf · problem set 6-1 1. all measurements...

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