CHAPTER EIGHT. Side opposite Hypotenuse = 24. sinθ = Side adjacent Hypotenuse = 10. cosθ = Side opposite Hypotenuse = 6. Side adjacent Hypotenuse = 9

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1 CHAPTER EIGHT 8.1 SOLUTIONS 483 Solutions for Section 8.1 Eercises 1. By the Pythgoren theorem, the hypotenuse hs length =. () tnθ = opposite djcent = 2 1 = 2. (b) sinθ = opposite hypotenuse = 2. (c) cosθ = djcent hypotenuse = By the Pythgoren theorem, the hypotenuse hs length = 12. () sinθ = 12 (b) sinφ = (c) cosθ = (d) cosφ = 12 (e) tnθ = 10 = 1 2 (f) tnφ = 10 = 2 3. () We hve (b) We hve 4. () We hve (b) We hve sinθ = cosθ = sinθ = cosθ = Side opposite Hypotenuse = = Side djcent Hypotenuse = = Side opposite Hypotenuse = = 0.. Side djcent Hypotenuse = = We use the Pythgoren theorem to find the length of the hypotenuse: Hypotenuse 2 = (0.1) 2 +(0.2) 2 = = 0.0 Hypotenuse = 0.0. () We hve (b) We hve sinθ = cosθ = Side opposite Hypotenuse = = Side djcent Hypotenuse = =

2 484 Chpter Eight /SOLUTIONS 6. We use the Pythgoren theorem to find the length of the opposite side: () We hve (b) We hve (Opposite side) 2 = (Opposite side) 2 = 1024 sinθ = cosθ = (Opposite side) 2 = 880 Opposite side = 880. Side opposite 880 Hypotenuse = = Side djcent Hypotenuse = = Since sin17 = y 7, we hve y = 7sin17. Similrly, since cos17 = 7, we hve = 7cos Since sin12 = 4 r, we hve r = 4 sin12. Similrly, since tn12 = 4, we hve = 4 tn Since cos37 = 6 r, we hve r = 6 cos37. Similrly, since tn37 = y 6, we hve y = 6tn Since sin40 = y 1, we hve y = 1sin40. Similrly, since cos40 = 1, we hve = 1cos Since tn77 = 9 y, we hve y = 9 tn77. Similrly, since sin77 = 9 r, we hve r = 9 sin Since sin22 = λ r, we hve r = λ sin22. Similrly, since tn22 = λ y, we hve y = λ tn We hve sinθ = 0.876, soθ = sin = We hve cosθ = 0.016, soθ = cos = We hve tnθ = 0.123, soθ = tn = We hve tnθ = 4.169, soθ = tn = We hve sinθ = 0.999, soθ = sin = We hve cosθ = 0.999, soθ = cos = We hve c = 2 +b 2 = sina = c A = sin 1 c = 3.38 B = 90 A = We hve b = c 2 2 = sina = c A = sin 1 c = 4.8 B = 90 A = We hve B = 90 A = 62 = c sina = 20sin28 = b = c sinb = 20sin62 =

3 8.1 SOLUTIONS We hve A = 90 B = 62 = c sina c = 20 sin62 = tnb = b 20 b = 20tn62 = Problems θ y P = (,y) = (cosθ,sinθ) Figure 8.1 In Section 7.2, the sine nd cosine re defined using the unit circle nd the coordintes of point P s cosθ = nd sinθ = y. Now we look t the tringle. Since the circle s rdius is 1, the hypotenuse of the tringle hs length 1. The sides of the tringle, nd y, re the coordintes of the point P. Therefore, the right tringle definitions give cosθ = Adjcent Hypotenuse = Opposite =. nd sinθ = 1 Hypotenuse = y 1 = y. Angles outside the intervl 0 θ 90 do not fit in right tringle, so the rgument only works for these vlues of θ. We conclude tht for these ngles, the two definitions give the sme result. 24. In right tringle, we hve sinθ = Opposite Hypotenuse Thus, rewriting nd cnceling, we hve nd cosθ = Adjcent Hypotenuse nd tnθ = Opposite Adjcent. sinθ cosθ = Opposite Hypotenuse Adjcent Hypotenuse = Opposite Hypotenuse Hypotenuse = Opposite Adjcent Adjcent = tnθ. 2. We hve c = = 2. Hence () sin4 = opposite hypotenuse = 2 = 1 2 = (b) cos4 = djcent hypotenuse = 2 = 1 2 = (c) tn4 = opposite djcent = =

4 486 Chpter Eight /SOLUTIONS 26. We hve PS = 1, by symmetry of tringle PQR. By the Pythgoren theorem, we hve = = 3. Using tringle P QS we hve () sin60 = opposite hypotenuse = 3 2. (b) cos60 = djcent hypotenuse = 1 2. (c) tn60 = opposite djcent = 3 1 = Figure 8.2 illustrtes this sitution. 200 h 30 Figure 8.2 We hve right tringle with legs nd200 nd hypotenuse h. Thus, sin30 = 200 h h = 200 sin30 = 200 = 400 feet. 0. To find the distnce, we cn relte the ngle nd its opposite nd djcent legs by writing tn30 = 200 = feet. tn30 We could lso write the eqution = h 2 nd substitute h = 400 ft to solve for. 28. See Figure Since the tngent is the length of the opposite side divided by the length of the djcent side, The width of the river is bout 80 meters. tn8 = d 0 d = 0tn Let the origin be t the rendezvous point nd the positive -is point est. Then, s she reches the river, the volunteer s coordintes re (,1.3), where is her distnce est of the rendezvous point, in miles. Since the slope of the line trced by the volunteer s pth is Slope = Δy Δ = = 1.3, nd the line forms 7 ngle with the positive -is, we hve: tn7 = 1.3 or = 1.3 = miles. tn7 Thus, the volunteer is locted bout 0.3 miles est of the rendezvous point s she reches the river.

5 8.1 SOLUTIONS Since (0, 0) nd (16, b) re points on the line, Since the line forms n ngle of 0, we lso hve The two epressions together give Slope of line = Δy Δ = b = b 16. Slope of line = tn0 b 16 = tn0 which mens b = 16tn0 = This nswer mkes sense becuse n ngle slightly bigger thn4 should give y-coordinte slightly lrger thn 16, the -coordinte. 31. Since (0, 0) nd (p, ) re points on the line, Since the line forms n ngle of 1 we lso hve The two epressions together give Slope of line = Δy Δ = 0 p 0 = p. Slope of line = tn(1 ) p = tn(1 ) which mens p = tn(1 ) = Since (0, 0) nd ( 3, q) re points on the line, Since the line forms n ngle of 3π 8, we lso hve Slope of line = Δy q 0 = Δ 3 0 = q 3. The two epressions together give Slope of line = tn 3π 8 q 3 = tn 3π 8 which mens q = 3tn 3π 8 = This nswer mkes sense becuse point with negtive -coordinte on line with positive slope should hve negtive y-coordinte. 33. Since (0, 0) nd (p, 1) re points on the line, Since the line forms n ngle of π 6, we lso hve The two epressions together give 1 p = tn π 6 Slope of line = Δy Δ = 1 0 p 0 = 1 p. Slope of line = tn π 6. which mens 1 p = tn(π 6) = This nswer mkes sense becuse point left of the y-is on line through the origin with negtive slope should be bove the -is.

6 488 Chpter Eight /SOLUTIONS 34. Using right tringle trigonometry, we hve tnθ = ( ) 240 θ = (tn) = Letd be the horizontl distnce from the irplne to the rch. See Figure 8.3. Then,tnθ = 3000 d, ord = 3000 tnθ feet. Plne θ 3,000 θ d Arch Figure Since the distnce from P toais 0 tn42 nd the distnce fromp tob is 0 tn3, d = 0 tn feet. tn () Since the grde of the rmp is7% = 7 100, this mens tht 7-foot height difference occurs over horizontl distnce of 100 feet. So we hve tnθ = 7. Using 100 thetn 1 button on the clcultor, we get ( ) 7 θ = tn 1 = θ y ft ft Figure ft (b) From the right tringle representing the rmp we see tht one leg represents the height difference between the drivewy nd the front door, which is 2 ft. The other leg represents the drivewy, of which we would like to find the length. Then 2 = tn Solving this eqution for, we hve = 2 tn = So the drivewy hs to be 28.7 feet long. (We cn lso use similr tringles: 2 7 = 100.) (c) The rmp is represented by the hypotenuse y of the right tringle. Using the Pythgoren theorem we hve y = = feet. (We cn lso use sin = 2 y.) 38. Since the rise, the run nd the roof form right tringle with horizontl leg of 12 nd verticl leg of 10, we hvetnθ = Using thetn 1 button on the clcultor, we hve θ = tn 1 ( ) =

7 8.1 SOLUTIONS y y = 2+ 1 θ Figure 8. They-intercept of this line is ; the-intercept is2.. These re the lengths of the legs of the right tringle in Figure 8. sotnθ = 2. = 2, or θ = tn 1 (2) () The side opposite of ngle φ hs lengthbnd the side djcent to ngle φ hs length. Therefore, (b) side opposite sinφ = hypotenuse = b c side djcent cosφ = hypotenuse = c side opposite tnφ = side djcent = b. side opposite φ sinφ = = b hypotenuse c, side djcent toθ cosθ = = b hypotenuse c. Thus sin φ = cos θ. Reversing the roles of φ nd θ, one cn show cos φ = sin θ in ectly the sme wy. 41. To solve for the distnce, we use tn3 = 94 nd solve for : = 94 tn3 = ft. To solve for the height of the Sefirst Tower, we cn use tn37 = y nd solve for y: y = tn37 = ft. (The ctul height of the Sefirst Tower is43 ft.) 42. See Figure 8.6. The ngle θ is the sun s ngle of elevtion. Here,tnθ = 0 60 = ( ) 6. So,θ = tn θ 60 Figure 8.6

8 490 Chpter Eight /SOLUTIONS 43. Drw picture s in Figure 8.7. The ngle tht we wnt is lbeled θ in this picture. We see tht tnθ = = Evlutingtn 1 (1.73) on clcultor, we get θ θ 10 Figure () First ssume A = 30 nd b = 2 3. Since this is right tringle, we know tht B = = 60. Now we cn determine by writing cosa = djcent hypotenuse = We lso know tht cosa = 2, becuse A is 30. So 2 = 2 3, which mens tht = 4. It follows from the Pythgoren theorem thtc is = 2. (b) Now ssume tht = 2 nd c = 24. The Pythgoren theorem, 2 +b 2 = c 2, implies b = = 7. To determine ngles, we use sina = opposite hypotenuse = c = 24 ( ) So evlutingsin 1 on the clcultor, we find 2 thta = Therefore,B = = (Be sure tht your clcultor is in degree mode when using the sin 1.) 4. In the figure, we see thttnα = 2 10 = 0.2 nd tnβ = (2+1) 10 = 0.3. Thus, tnα = 0.2, so α = tn 1 (0.2) tnβ = 0.3, so β = tn 1 (0.3) Let d be the distnce from the bse of the ldder to the wll; see Figure 8.8. Then, d 3 = cosα, sod = 3cosα meters. 3 α d Figure () Since sin4 = h 12, we hve h = 12sin feet. (b) Since sin30 = h 12, we hve h = 12sin30 = 62. feet. (c) Since cos4 = c 12, we hve c = 12cos(4 ) feet. Since cos30 = d 12, we hve d feet.

9 8.2 SOLUTIONS 491 Solutions for Section 8.2 Eercises 1. By the Lw of Sines, we hve sin100 = 6 sin18 ( sin100 ) = 6 sin By the Lw of Cosines, we hve 2 = (3)()cos(21 ) In Figure 8.9, we hve β = β = 2. sin38 = 4 c 4 c = sin c 2 = 4 2 +b 2 b = c c β 4 38 b Figure In Figure 8.10, we hve tnθ = 7 2 = 3. θ = tn 1 (3.) θ tnψ = 2 7 ψ = tn 1 ( 2 7 ) ψ c 2 = = 3 c =

10 492 Chpter Eight /SOLUTIONS ψ c 7 θ 2 Figure In Figure 8.11, we hve θ = θ = 80. = 12cos10 12(0.98) b = 12sin10 b 12(0.174) b θ b Figure Using the Lw of Cosines in Figure 8.12, we hve b 2 = cos32 b.979. Using the Lw of Cosines gin to findθ, 11 2 = cosθ cosθ = θ ψ = θ 8 Figure 8.12 ψ b

11 8.2 SOLUTIONS Using the Lw of Sines in Figure 8.13, we hve sinθ = sin20 6 θ = sin 1 ( sin20 6 ) = This is correct since θ < 90 in the tringle. We epect θ < 20 becuse θ is opposite side which is shorter thn 6. Therefore ψ = = sin = sin20 6 = 6sin sin20 = θ 6 ψ 20 Figure In Figure 8.14, we first determine α using the Lw of Cosines: 7 2 = cosα 240cosα = 19 cosα = , α = cos 1 ( ) γ 7 α 10 Figure 8.14 β Then, we determine β using the Lw of Cosines: Finlly, γ = 180 α β = cosβ cosβ = 140 β = cos 1 ( 140 )

12 494 Chpter Eight /SOLUTIONS 9. Using the Lw of Cosines, we hve 41 2 = cosC 497 = 1120cosC ( C = cos ) 1120 C = Using the Lw of Cosines gin (though we could use the Lw of Sines), we hve Thus, B = 180 A C = The Lw of Cosines gives 20 2 = cosA 206 = 2296cosA ( ) 206 A = cos A = c 2 = cos23 = c =.417 From the Lw of Sines, we cn solve for B, which must be n cute ngle since it is cross from b, which is shorter thn, sob < A, nd there cn be no more thn one obtuse ngle in tringle. soa = 180 B C = Using the Lw of Cosines, we hve Using the Lw of Sines, we hve sinb 12 = sin sinb = 12sin(23 ).417 ( ) 12sin(23 B = sin 1 ).417 = 7.789, b 2 = cos81 b 2 = b = sina 20 = sin sina = A = sin 1 (0.622) or 180 sin 1 (0.622) A = or We cn discrd the lrger nswer, s it does not fit in tringle withb = 81. Thus we hvec = 180 A B = Using the Lw of Cosines, we hve c 2 = cos12 = c =

13 8.2 SOLUTIONS 49 Using the Lw of Sines, we hve sina 20 = sin sina = 20 sin ( ) ( ) A = sin 1 20 sin12 or 180 sin 1 20 sin A = or Since B is lrger ngle thn A (since b > ), we cn discrd the lrger nswer, s it does not fit in tringle. Thus we hve B = 180 A C = We begin by using the Lw of Cosines to find side c: c 2 = cos80 c 2 = c = We cn now use the Lw of Sines to find the other two ngles. Therefore, A = = By the Lw of Cosines, sinb 8 = sin sinb = 8 sin B = sin B = c 2 = cos114 c 2 = c = By the Lw of Sines, sina = sin A = Tht mens B = 180 A C = By the Lw of Cosines, c 2 = cos32 c = Angle A is cute (since < b nd thus A < B), so we solve for it using the Lw of Sines. We hve This gives B = 180 A C = sina sina = sinc c = sin sina = sin A = sin 1( sin32 ) =

14 496 Chpter Eight /SOLUTIONS 16. We begin by finding the ngle C, which is = 142. We cn now use the lw of sines to find the other two sides. Similrly, b sin2 = 4 sin142 b = sin2 b = sin13 = 4 sin142 = sin13 = sin142 4 sin From the Lw of Sines, we hve sina sina 8 = sinc c = sin98 17 ( A = sin 1 8sin98 17 = Since C is obtuse, A nd B re cute, so this is the correct vlue of A. (Hd A been obtuse, we would hve hd to subtrct this vlue from 180 to compenste.) Now we hve B = 180 A C = Once gin, we cn use the Lw of Sines to solve forb: b sinb = c sinc b sin4.224 = 17 sin98 ) b = 17sin4.224 sin98 = We begin by finding the ngle B, which is = 18. We cn now use the lw of sines to find the other two sides. Similrly, sin12 = sin10 = sin12 = b sin18 = sin10 b = sin18 b = sin10 sin10

15 8.2 SOLUTIONS We begin by finding the ngle B, which is = 3. We cn now use the lw of sines to find the other two sides. Similrly, sin92 = 9 sin3 = sin92 = b sin3 = 9 sin3 b = sin3 b = sin3 9 sin3 20. We begin by finding the ngle B, which is180 9 = 166. We cn now use the lw of sines to find the other two sides. Similrly, sin = 3 sin9 = sin = b sin166 = 3 sin9 b = sin166 b = sin9 3 sin9 21. First, we recognize tht it is possible tht there re two tringles, since we my hve the mbiguous cse. However, since 9 is greter thn90, there re no other obtuse ngles possible, so there is but one possible tringle. We begin by finding the ngle C using the Lw of Sines: sinc 10 = sin9 sinc = 10 sin9 C = sin Since there is no rcsine of 1.992, we notice tht there is problem. There re no solutions. We could hve seen this before becuse the longest side is lwys cross from the lrgest ngle, nd since C cnnot be greter thn 9, the side cross from it (10) cnnot be longer thn the side cross from 9. Since it is bigger, no tringle fulfills the conditions given. 22. First, we recognize tht it is possible tht there re two tringles, since we my hve the mbiguous cse. However, we know tht ngle C must be less thn 72, since the side cross from it is shorter thn 13. Thus, we begin by finding the ngle C using the lw of sines: sinc 4 = sin72 13 sinc = 4 sin72 13 C = sin C =

16 498 Chpter Eight /SOLUTIONS We cn now solve for A, which is = Using the lw of sines, we cn solve for side : sin = 13 sin72 = sin = sin First, we recognize tht it is possible tht there re two tringles, since we my hve the mbiguous cse. However, we know tht nglec must be less thn7, since the side cross from it is shorter thn 7. Thus, we begin by finding the ngle C using the lw of sines: sinc 2 = sin7 7 sinc = 2 sin7 7 C = sin C = We cn now solve for A, which is = Using the lw of sines, we cn solve for side : sin = 7 sin7 = sin = sin7 24. First, we recognize tht it is possible tht there re two tringles, since we my hve the mbiguous cse. Therefore, we know tht ngle C might be obtuse or cute. We tke this into ccount while using the lw of sines to findc: sinc 8 = sin17 sinc = 8 sin17 C = sin C = However, C could lso equl 180 sin = Thus, A cn be either = or = Using the lw of sines, we cn solve for side in either cse: or sin = sin17 = sin = , sin = sin17 = sin = sin17 sin17

17 8.2 SOLUTIONS We begin by finding the ngle C, which is π 3π π 20π 12π π = We cn now use the Lw of Sines to find the other two sides. b sin(π 20) = 1 sin(7π 20) Similrly, b = sin(π 20) b = = 7π 20 rdins. 1 sin(7π 20) sin(3π ) = 1 sin(7π 20) 1 = sin(3π ) sin(7π 20) = We begin by finding the ngle C, which is π 2π 3 π 9π 6π π = 9 9 We cn now use the lw of sines to find the other two sides. = 2π 9 rdins. Similrly, b sin(π 9) = 7 sin(2π 9) 7 b = sin(π 9) sin(2π 9) b = sin(2π 3) = 7 sin(2π 9) = sin(2π 3) = sin(2π 9) 27. We begin by finding the ngle C, which is π 3π 8 π 8π 3π 2π = 4 8 We cn now use the lw of sines to find the other two sides. = 3π 8 rdins. Similrly, b sin(π 4) = 9 sin(3π 8) 9 b = sin(π 4) sin(3π 8) b = sin(3π 8) = 9 sin(3π 8) = 9. Note tht this tringle hs two equl ngles, nd therefore two equl sides: it is n isosceles tringle.

18 00 Chpter Eight /SOLUTIONS 28. We use the Lw of Cosines in Figure 8.1 to find the length of side b, getting Sob cm. To findγ we use the Lw of Cosines: Then α = = b 2 = (21)(18.7)cos = cosγ cosγ = γ = cos 1 ( ) = γ α b Figure b 2.8 γ c Figure 8.16 b The Lw of Sines tells us tht = 2, so b = m. We hve γ = sin10. sin = We use this vlue to find side lengthc. We hve c 2 = 2 2 +(.837) 2 2(2)(.837)cos , or c = m. 30. In Figure 8.17, use the Lw of Sines: sin(30 ) 29 = sinβ 10 to obtinsinβ nd usesin 1 to findβ orβ We then knowα 1 = , or α 2 = We cn use the vlue of α nd the Lw of Sines to find the length of side : 1 sin( ) = 29 sin30, 2 or sin( ) = 29 sin ft ft α 1 α γ β 1 γ β Figure 8.17

19 8.2 SOLUTIONS We use the Lw of Cosines in Figure 8.18 to findα nd β: 16 2 = (20)(24)cosα cosα = 0.7 α = (20)(16)cos(β) cosβ = 0.12 β Problems Finllyγ = = β 20 γ 24 Figure () In right tringle sin θ = Opp Hyp. Thus sin θ = 3 7. To findsinφ we use the Lw of Sines: sinφ 1 = sin(20 ). 8 This implies tht sinφ = 1sin(20 ). 8 (b) Since, using the definition of the sine,sinθ = 3 7, θ = sin 1( ) This mkes sense, s we epect 0 < θ < 90. Forsinφ = 1sin(20 ), using the Lw of Sines, there re two solutions: 8 ) φ = sin 1 ( 1sin(20 ) 8 α ( ) 1sin( nd φ = 180 sin 1 ) Assuming thtφ > 90, s the figure suggests, we choose the second solution, φ The method for finding θ uses the definition of the sine, since θ is n ngle in right tringle. Becuse φ is n ngle in non-right tringle we find it using the Lw of Sines, nd we must decide whether to use the cute or obtuse solution. 33. () By the Lw of Sines, we hve sinθ = sin ( ) 3 sinθ = sin (b) If sinθ = 0.282, then θ (s found on clcultor) or θ Since the tringle lredy hs 110 ngle, θ (The ngle would be too lrge.) (c) The height of the tringle is 10sinθ = = cm. Since the sum of the ngles of tringle is 180, nd we know two of the ngles,θ = nd110, so the third ngle is = By the Lw of Sines, we hve Thus, the tringle hs sin = sin3.626 Bse Bse = 10sin3.626 sin cm. Are = 1 2 Bse Height = = cm2.

20 02 Chpter Eight /SOLUTIONS 34. In Figure 8.19, the fire sttions re tandbnd the forest fire is tc. The ngle tc is = 68. Solving for nd b using the Lw of Sines, we get 6.7 sin68 = 6.7 sin4 sin68 = b sin8 = 6.7sin4 b = 6.7sin8 sin68 sin68 = b = The fire sttion t point B is closer by = miles. C b 68 A miles Figure 8.19 B 3. Figure 8.20 shows the Eiffel Tower, CD, nd the two ngles of elevtion, CAD = 60 nd CBD = 70. In tringle ABC, we find ABC = = 110 nd ACB = = 10. We use the Lw of Sines to solve for : 210 sin10 = sin60 = 210sin60 = sin10 Using the vlue of in the right tringle CBD to findy, we get sin70 y = y = sin70 ( ) feet. 10 C y A B 70 D Figure 8.20

21 8.2 SOLUTIONS After hlf n hour, the A380 hs trveled 09 (1 2) = 24. miles nd the 787 hs trveled 03 (1 2) = 21. miles. See Figure Using the Lw of Cosines: 2 = (24.)(21.)cos103 = miles. A Kennedy Figure Figure 8.22 shows the tringle; we wnt to find. The other two ngles re(180 39) 2 = 70.. Using the Lw of Sines: 42 sin39 = sin70. = 42sin70. sin39 = feet ft Figure The horizontl nd verticl displcements of the imge re given by Δ = 12cos2 = Δy = 12sin2 =.071. Thus, the new coordintes re (,y) = (8+Δ,+Δy) = (18.876,10.071). 39. Initilly, the ngle θ mde by the imge with respect to the origin is given by nd the distnce from the origin is given by tnθ = 3 θ = tn 1 (3 ) , r = = 34. After the rottion, the new ngle isφ = θ+42 = , but the distnce does not chnge. The- ndy-coordintes re = rcosφ = 34cos( ) = y = rsinφ = 34sin( ) =.7, so the new position is (1.708,.7).

22 04 Chpter Eight /SOLUTIONS 40. If we look t Figure 8.23, we see tht there re two tringles: the originl tringle with ngles A, B, C nd the right tringle with hypotenuse b. A h b c θ C Figure 8.23 B or The Pythgoren theorem gives 2 +h 2 = b 2, h 2 = b 2 2. If we pply the Pythgoren theorem to the right tringle with legs h nd + we obtin Substituting h 2 = b 2 2 into this eqution gives This, in turn, reduces to (+) 2 +h 2 = c b 2 2 h 2 2 +b 2 +2 = c 2. We now determine. SinceC is obtuse, cosc will be negtive. We hve or cosc = cosθ = b, = bcosc. = c 2. Substituting this epression forinto our eqution gives the Lw of Cosines: 2 +b 2 2bcosC = c From the figure, we see thtsina = h b, which gives h = bsina. We lso hve sinb = h, which gives h = sinb. Thus, b sin A = sin B, which gives the Lw of Sines: sina = sinb b. 42. () See Figure Letbe the distnce from the pitcher s rubber to first bse. Then, by the Lw of Cosines, 2 = (60.)(90)cos4 = To find the distnce from the pitcher s rubber to second bse, let y be the distnce from home plte to second bse. Then y 2 = y =

23 8.2 SOLUTIONS 0 Then we find the distnce from the pitcher s rubber to second: Distnce = = From the pitcher s rubber to first bse is closer by = feet. y Third bse z Home plte Pitcher s rubber Second bse 30 Bll in prt (b) First bse Figure 8.24 (b) Using result from prt (), the distnce from 30 feet pst second bse to home plte is given by Distnce = = feet. Let z be the distnce from 30 feet pst second bse to third bse: z 2 = (30)(90)cos13 z = feet. 43. From Figure 8.2, we see ABC = = 38. Using the Lw of Sines, we hve 102 sin38 = c sin49 c = 102sin49 sin38 = feet. B 38 c A ft Figure 8.2 C

24 06 Chpter Eight /SOLUTIONS 44. The three stkes re t A,B,C in Figure Using the Lw of Cosines, we hve 2 = (82)(97)cos The mound is pproimtely feet wide. A Mound 82 ft ft B C Figure One wy to orgnize this sitution is to use the bbrevitions from high school geometry. The si possibilities re {SSS,SAS,SSA,ASA,AAS,AAA}. SSS Knowing ll three sides llows us to find the ngles by using the Lw of Cosines. SAS Knowing two sides nd the included ngle llows us to find the third side length by using the Lw of Cosines. We cn then use the SSS procedure. SSA Knowing two sides but not the included ngle is clled the mbiguous cse, becuse there could be two different solutions. Use the Lw of Sines to find one of the missing ngles, which, becuse we use the rcsine, my give two vlues. Or, use the Lw of Cosines, which produces qudrtic eqution tht my lso give two vlues. Treting these cses seprtely, we cn continue to find ll sides nd ngles using the SAS procedure. ASA Knowing two ngles llows us to esily find the third ngle. Use the Lw of Sines to find ech side. AAS Find the third ngle nd then use the Lw of Sines to find ech side. AAA This hs n infinite number of solutions becuse of similrity of tringles. Once one side is known, then the ASA or AAS procedure cn be followed. 46. Using the Lw of Sines on tringle DEF in Figure 8.27, we hve 10.2 sin29 = ED sin68 ED = 10.2sin68 sin29 = feet. Totl mount of wire needed = = feet. Since wire is sold in 100-foot rolls, 4 rolls of wire re needed. F 68 House 10.2 ft ft E Figure ft D Pole 47. () By the Lw of Sines, sin = sinψ 112.

25 8.2 SOLUTIONS 07 Solving for ψ gives 112 sin = sinψ ( ψ = sin sin ) Thus θ = Knowing the ngle θ now llows us to solve for the distnce LT. By the Lw of Cosines: LT = (112)(43)cos ft. This nswer cn be lso seen by figuring out thttol is n isosceles tringle nd thereforelt = TO = 43 ft. (b) Since the lot is n isosceles tringle, the height,h, from pointt to the bseol splits the lot into two congruent right tringles. We then hve: giving This is bout hlf n cre. tn82.6 = h () Using the Lw of Sines in Figure 8.4, we hve (b) From Figure 8.46, we get or h = 6tn82.6, Are = 1 2 (112)(6tn82.6 ) 24,14.86 ft 2. sinθ 8 = sin40 7 sinθ = 8 7 sin40 = 0.73 θ = sin 1 (0.73) = sinφ 8 = sin40 7 sinφ = 8 7 sin40 = This is the sme eqution we hd forθ in prt (). However, judging from the figures,φis not equl toθ. Knowing the sine of n ngle is not enough to tell us the ngle. In fct, there re two ngles between0 nd 180 whose sine is One of them isθ = sin 1 (0.73) = Figure 8.28 shows the other isφ = 180 θ = y Figure 8.28

26 08 Chpter Eight /SOLUTIONS 49. 2π 1 π b 8 2π 3 h Figure 8.29 () sin(2π 3) 8 = sin(π ), so = 8sin(π ) 8sin(2π 1) =.420. Similrly,b = = sin(2π 3) sin(2π 3) (b) Construct n ltitude h s in Figure We hve sin(π ) = h, so h = 8sin(π ) = Then re of the 8 tringle is 1 (3.77)(4.702) = Solutions for Section 8.3 Eercises 1. Qudrnt IV. 2. Qudrnt I. 3. Since 470 = , such point is in Qudrnt II. 4. Since 2.4π = 2π +0.4π, such point is in Qudrnt I.. Since 3.2π = 2π +1.2π, such point is in Qudrnt III. 6. Since 2.9π = 2 2π + 1.1π, such point is in Qudrnt III. 7. Since 10.3π = 2π + 0.3π, such point is in Qudrnt I. 8. Since 13.4π = 7 2π +0.6π, such point is in Qudrnt II. 9. Since 7 is n ngle of 7 rdins, corresponding to rottion of just over 2π, or one full revolution, in the clockwise direction, such point is in Qudrnt IV. 10. We hve 0 < θ < 90. See Figure y (2,3) θ Figure 8.30

27 8.3 SOLUTIONS We hve 90 < θ < 180. See Figure ( 1, 7) y θ Figure We hve 270 < θ < 360. See Figure y θ (2, 3) Figure We hve 180 < θ < 270. See Figure θ y ( 1.1, 3.2) Figure With = 1 nd y = 1, find r from r = 2 +y 2 = = 2. Find θ from tnθ = y = 1 1 = 1. Thus, θ = tn 1 (1) = π 4. Since (1,1) is in the first qudrnt this is correct θ. The polr coordintes re( 2,π 4). 1. With = 1 nd y = 0, find r = 2 +y 2 = ( 1) = 1. Find θ from tnθ = y = 0 ( 1) = 0. Thus, θ = tn 1 (0) = 0. Since ( 1,0) is on the -is between the second nd third qudrnt, θ = π. The polr coordintes re (1,π). 16. With = 6 nd y = 2, find r = ( 6) 2 +( 2) 2 = 8 = 2 2. Find θ from tnθ = y = 2 6 = 1 3. Thus, θ = tn 1 ( 1 3) = π 6. Since ( 6, 2) is in the fourth qudrnt, this is the correct θ. The polr coordintes re (2 2, π 6). 17. With = 3 nd y = 1, find r = ( 3) = 4 = 2. Find θ from tnθ = y = 1 ( 3). Thus, θ = tn 1 ( 1 3) = π 6. Since ( 3,1) is in the second qudrnt, θ = π 6 + π = π 6. The polr coordintes re (2,π 6).

28 10 Chpter Eight /SOLUTIONS 18. With = 3 nd y = 10, we find r = 2 +y 2 = ( 3) = 109. Find θ from tnθ = y = 10 3 which gives θ = tn 1 (10 3) = Since ( 3,10) is in the second qudrnt nd θ is in the fourth qudrnt, the correct ngle isθ +π = The polr coordintes re ( 109,1.8622). 19. Withr = 1 nd θ = 2π 3, we find = rcosθ = 1 cos(2π 3) = 1 2 nd y = rsinθ = 1 sin(2π 3) = 3 2. The rectngulr coordintes re ( 1 2, 3 2). 20. With r = 3 nd θ = 3π 4, we find = rcosθ = 3cos( 3π 4) = 3( 2 2) = 6 2 nd y = rsinθ = 3sin( 3π 4) = 3( 2 2) = 6 2. The rectngulr coordintes re ( 6 2, 6 2). 21. With r = 2 3 nd θ = π 6, we find = rcosθ = 2 3cos( π 6) = = 3 nd y = rsinθ = 2 3sin( π 6) = 2 3( 1 2) = 3. The rectngulr coordintes re (3, 3). 22. With r = 2 nd θ = π 6, we find = rcosθ = 2cos(π 6) = 2( 3 2) = 3 nd y = rsinθ = 2sin(π 6) = 2(1 2) = 1. The rectngulr coordintes re ( 3,1). Problems 23. Since r = 2 +y 2, write 2 +y 2 = 2. Squring both sides gives 2 +y 2 = Multiply both sides byrto trnsform the right side of the eqution into6rcosθ so tht we cn substitute = rcosθ. The left side is now r 2 = 2 +y 2. In rectngulr coordintes, the eqution is 2 +y 2 = Sinceθ = tn 1 (y ), writetn 1 (y ) = π. Tking the tngent of both sides, we gety = tn(π 4) = 1. In rectngulr 4 coordintes, the eqution is y =. 26. Rewrite the left side using tnθ = sinθ cosθ : sinθ = rcosθ 2 cosθ In order to substitute = r cos θ nd y = r sin θ, multiply by the r in the numertor nd denomintor on the left nd substitute: So or rsinθ = rcosθ 2 rcosθ y = 2. y = ( 2) y = By substitution the eqution becomes 3r cos θ 4r sin θ = 2. This cn be written s r = 2 3cosθ 4sinθ. 28. By substituting r 2 = 2 +y 2, we hve r 2 =. Since r is positive, this could lso be written s r =. 29. By substituting = rcosθ nd y = rsinθ, the eqution becomes rsinθ = (rcosθ) 2. This could lso be written s r = sinθ cos 2 θ. 30. By substituting = rcosθ ndy = rsinθ, the eqution becomes 2rcosθ rsinθ = 1. This cn be written s r = 1 2cosθsinθ.

29 8.3 SOLUTIONS Figure 8.34 shows tht t 12 noon, we hve: In Crtesin coordintes, H = (0,3). In polr coordintes, H = (3,π 2); tht isr = 3,θ = π 2. In Crtesin coordintes, M = (0,4). In polr coordintes, M = (4,π 2), tht is r = 4,θ = π 2. M = (0,4) H = (0,3) 90 Figure Figure 8.3 shows tht t 3 pm, we hve: In Crtesin coordintes, H = (3,0). In polr coordintes, H = (3,0); tht is r = 3,θ = 0. In Crtesin coordintes, M = (0,4). In polr coordintes, M = (4,π 2), tht is r = 4,θ = π 2. M = (0,4) 90 H = (3,0) Figure Figure 8.36 shows tht t 9 m, we hve: In Crtesin coordintes, H = ( 3,0). In polr coordintes, H = (3,π); tht is r = 3,θ = π. In Crtesin coordintes, M = (0,4). In polr coordintes, M = (4,π 2), tht is r = 4,θ = π 2. H = ( 3,0) M = (0,4) θ = π Figure Figure 8.37 shows tht t 1 pm, we hve: In Crtesin coordintes, M = (0, 4). In polr coordintes, M = (4, π 2); tht is r = 4, θ = π 2. At 1 pm, the hour hnd points towrd 1, so forh, we hve r = 3 ndθ = π 3. Thus, the Crtesin coordintes ofh re ( ( π π = 3cos = 1., y = 3sin = 3) 3) 3 3 =

30 12 Chpter Eight /SOLUTIONS M = (0,4) π 3 H = (1.,2.98) Figure Figure 8.38 shows tht t 1:30 pm, the polr coordintes of the point H (hlfwy between 1 nd 2 on the clock fce) re r = 3 ndθ = 4 = π 4. Thus, the Crtesin coordintes of H re given by ( π = 3cos = 4) 3 2 ( π 2.121, y = 3sin = 2 4) In Crtesin coordintes, H (2.121, 2.121). In polr coordintes, H = (3, π 4). In Crtesin coordintes, M = (0, 4). In polr coordintes, M = (4, 3π 2). H = (2.12,2.12) 4 M = (0, 4) Figure Figure 8.39 shows tht t 7 m the polr coordintes of the point H re r = 3 nd θ = = 240 = 4π 3. Thus, the Crtesin coordintes ofh re given by ( ) 4π = 3cos = 3 ( ) 4π 3 2 = 1., y = 3sin = Thus, in Crtesin coordintes, H = ( 1., 2.98). In polr coordintes, H = (3, 4π 3). In Crtesin coordintes, M = (0, 4). In polr coordintes, M = (4, π 2). θ 60 M = (0,4) H = ( 1., 2.60) Figure 8.39

31 8.3 SOLUTIONS Figure 8.40 shows tht t 3:30 pm, the polr coordintes of the point H (hlfwy between 3 nd 4 on the clock fce) re r = 3 ndθ = = 23π 12. Thus, the Crtesin coordintes of H re given by ( ) ( ) 23π 23π = 3cos 2.898, y = 3sin We hve: In Crtesin coordintes, H (2.898, 0.776); M = (0, 4). In polr coordintes, H = (3, 23π 12); M = (4, 3π 2). θ 7 H = (2.90, 0.78) M = (0, 4) Figure Figure 8.41 shows tht t 9:1 m, the polr coordintes of the point H (hlfwy between 9 nd 9:30 on the clock fce) re r = 3 nd θ = = 172.π 180. Thus, the Crtesin coordintes of H re given by ( ) ( ) 172.π 172.π = 3cos 2.974, y = 3sin In Crtesin coordintes, H ( 2.974, 0.392). In polr coordintes, H = (3, 172.π 180). In Crtesin coordintes, M = (4,0). In polr coordintes, M = (4,0). H = ( 2.79,0.39) 82. θ M = (4,0) Figure The region is given by 8 r 18 nd π 4 θ π The region is given by0 r 2 nd π 6 θ π The circulr rc hs equtionr = 1, for0 θ π 2. The verticl line = 2 hs polr equtionrcosθ = 2, orr = 2 cosθ. So the region is described by0 θ π 2 nd 1 r 2 cosθ. 42. () Tble 8.1 contins vlues of r = 1 sin θ, both ect nd rounded to one deciml. Tble 8.1 θ 0 π 3 π 2 2π 3 π 4π 3 3π 2 π 3 2π 7π 3 π 2 8π 3 r r

32 14 Chpter Eight /SOLUTIONS (b) See Figure y y Figure 8.42 Figure 8.43 (c) The circle hs eqution r = 1 2. The crdioid is r = 1 sin θ. Solving these two simultneously gives or 1 2 = 1 sinθ, sinθ = 1 2. Thus, θ = π 6 or π 6. This gives the points (,y) = ((1 2)cosπ 6,(1 2)sinπ 6) = ( 3 4,1 4) nd (,y) = ((1 2)cosπ 6,(1 2)sinπ 6) = ( 3 4,1 4) s the loction of intersection. (d) The curver = 1 sin2θ, pictured in Figure 8.43, hs two regions insted of the one region thtr = 1 sinθ hs. This is becuse 1 sin2θ will be 0 twice for every2π cycle inθ, s opposed to once for every2π cycle inθ for 1 sinθ. 43. There will be n loops. See Figures Figure 8.44: n = 1 Figure 8.4: n = 2 Figure 8.46: n = 3 Figure 8.47: n = The grph will begin to drw over itself for ny θ 2π so the grph will look the sme in ll three cses. See Figure Figure 8.48

33 8.3 SOLUTIONS 1 4. The curve will be smller loop inside lrger loop with n intersection point t the origin. Lrger n vlues increse the size of the loops. See Figures Figure 8.49: n = Figure 8.0: n = Figure 8.1: n = See Figures 8.2 nd 8.3. The first curve will be similr to the second curve, ecept the crdioid (hert) will be rotted clockwise by90 (π 2 rdins). This mkes sense becuse of the identitysinθ = cos(θ π 2). Figure 8.2: r = 1 cosθ Figure 8.3: r = 1 sinθ 47. Let 0 θ 2π nd 3 16 r A loop strts nd ends t the origin, tht is, when r = 0. This hppens first when θ = π 4 nd net when θ = π 4. This cn lso be seen by using trce mode on clcultor. Thus restricting θ so tht π 4 θ π 4 will grph the upper loop only. See Figure 8.4. To show only the other loop use 0 θ π 4 nd π 4 θ 2π. See Figure 8.. Figure 8.4: π 4 θ π 4 Figure 8.: 0 θ π 4 nd π 4 θ 2π 49. () Let 0 θ π 4 nd 0 r 1. (b) Brek the region into two pieces: one with nd 0 y, the other with nd 0 y 1 2.

34 16 Chpter Eight /SOLUTIONS 0. () The ngle directly north of the tower is θ = π 2. Thus the point on the crdioid t θ = π 2 hs rdius r = 1(1 + cosπ 2) = 1, so it is locted 1 miles north of the tower. (b) The ngle directly est of the tower is t ngleθ = 0. We hver = 1(1+cos0) = 30. Therefore, the polr coordintes of the point re (30,0). (c) We hve to find the polr coordintes of the house first. In Crtesin coordintes the house is locted t(2, 14). Thus we hve r = (2) = 72 = Also, cosθ = =.873, soθ = cos 1 (.873) =.10. Substituting this ngle into the eqution for the crdioid, we get r = 1(1+cos(.10)) = Therefore the house is miles wy from the tower nd lies just outside of the coverge re of the rdio signl. 1. We find the r vlues of the points on the curve t ngles π, 3π 4, nd 3π 4. They re 9.870,.2 nd.2, respectively. This mens tht t these ngles, the guests cn be no frther wy from the microphone if they wnt to be herd. The two guests t ngles±3π 4 hve to move closer to the microphone; for emple, they could sit t the sme ngles but only feet from the microphone. Solutions for Chpter 8 Review Eercises 1. By the Pythgoren theorem, we know tht the third side must be = 4. () Since sinθ is opposite side over hypotenuse, we hve sinθ = 4 7. (b) Since cos θ is djcent side over hypotenuse, we hve cos θ = 2 7. (c) Since tnθ is opposite side over djcent side, we hve tnθ = By the Pythgoren theorem, we know tht the third side must be = 106. () Since sinθ is opposite side over hypotenuse, we hve sinθ = 106. (b) Since cosθ is djcent side over hypotenuse, we hve cosθ = (c) Since tn θ is opposite side over djcent side, we hve tn θ = By the Pythgoren theorem, we know tht the third side must be = 80. () Since sin θ is opposite side over hypotenuse, we hve sin θ = (b) Since cosθ is djcent side over hypotenuse, we hve cosθ = (c) Since tnθ is opposite side over djcent side, we hve tnθ = Becuse the two ngles re the sme, we know tht the two missing sides must be equl. Therefore, by the Pythgoren theorem, we know tht the missing sides re (which we cll s) re given by: 17 2 = s 2 +s = 2s = s. () Since sinθ is opposite side over hypotenuse, we hve sinθ = = 1 2. (b) Since cosθ is djcent side over hypotenuse, we hve cosθ = = 1 2. (c) Since tn θ is opposite side over djcent side, we hve tn θ = 1.

35 . By the Pythgoren theorem, we know tht the third side must be = 117. () Since sinθ is opposite side over hypotenuse, we hve sinθ = (b) Since cos θ is djcent side over hypotenuse, we hve cos θ = (c) Since tnθ is opposite side over djcent side, we hve tnθ = By the Pythgoren theorem, we know tht the third side must be 2 +b 2. () Since sinθ is opposite side over hypotenuse, we hve sinθ = 2 +b 2. (b) Since cosθ is djcent side over hypotenuse, we hve cosθ = b 2 +b 2. (c) Since tn θ is opposite side over djcent side, we hve tn θ = b. 7. Since we re looking for the ngle θ, we hve tnθ = 0.999, soθ = tn = Since we re looking for the ngle θ, we hve sinθ = 3 ), soθ = sin 1( 3 = Since we re looking for the ngle θ, we hve tnθ =, soθ = 3 tn 1 = Since the output of the cosine function is 0 < θ < 1, there is no ngle whose cosine is We know tht cos30 = 2. Therefore, θ = We know tht sin30 = 1 2. Therefore,θ = We know tht cos4 = 2. Therefore, θ = Whenθ = 4, both the sine nd cosine functions hve the sme vlue. Therefore,θ = Since 299 is lmost 300, clockwise, such point is in Qudrnt I. 16. Since 730 = , such point is in Qudrnt I. 17. Since 7.7π = 3 2π 1.7π, such point is in Qudrnt I. 18. Since 14.4π = 7 2π + 0.4π, such point is in Qudrnt I. SOLUTIONS to Review Problems For Chpter Eight Withr = π 2 ndθ = 0, we find = rcosθ = (π 2)cos0 = 1.71 nd y = rsinθ = (π 2)sin0 = 0. The rectngulr coordintes re (1.71,0). 20. Withr = 2 nd θ = 2, we find = rcosθ = 2cos2 = ndy = rsinθ = 2sin2 = The rectngulr coordintes re ( 0.832,1.819). 21. Withr = 0, the point specified is the origin, no mtter wht the ngle mesure. So = rcosθ = 0 nd y = rsinθ = 0. The rectngulr coordintes re (0,0). 22. With r = 3 nd θ = 40, use clcultor in degree mode to find = rcosθ = 3cos(40 ) = nd y = rsinθ = 3sin(40 ) = The rectngulr coordintes re (2.298,1.928). Problems 23. We know ll three sides of this tringle, but only one of its ngles. We find the vlue ofsinθ ndsinφ in this right tringle: nd sinθ = opposite hypotenuse = 3 = 0.6 sinφ = opposite hypotenuse = 4 = 0.8. Using inverse sines, we know tht if sinφ = 0.8, then φ = sin 1 (0.8) Similrly sinθ = 0.6 mens θ = sin 1 (0.6) Notice φ+θ = 90, which hs to be true in right tringle.

36 18 Chpter Eight /SOLUTIONS 24. The other ngle must be θ = 90 9 = 31. (See Figure 8.6.) By definition of the tngent, tn9 = By definition of the sine, = tn sin9 = y y = sin y 9 θ Figure See Figure 8.7. By the Pythgoren theorem, = =. Thus φ = 90 θ sinθ = ( ) θ = sin θ θ φ 12 Figure The other ngle must be θ = = 10. (See Figure 8.8.) By the Lw of Sines, Agin using the Lw of Sines, y sin42 = 8 sin33 ( sin42 ) y = 8 sin sin10 = 8 sin33 ( sin10 ) = sin33

37 SOLUTIONS to Review Problems For Chpter Eight y θ 42 8 Figure Using tn13 = height 200 to find the height we get Usingcos13 = 200 to find the incline we get incline height = 200tn feet. incline = 200 cos feet. 28. First check to see if there is right tringle. It is not becuse So we must use the Lw of Cosines: = cosθ 2(63)(2) cosθ rccos(0.9231) θ θ 2 2 = (63)(2)cosθ 29. We hve tn28 = so Leg opposite = Leg djcent 4 4 = 4tn28 = 4(0.317) using clcultor = miles. 30. The ngle of observtion is lbeled θ in Figure 8.6. The distnce 200 meters forms the djcent side for this ngle nd the height of the blloon, h, is the opposite side. Thus, so tnθ = h 200, h = 200tnθ. 31. Consider lines OP 1 nd OP 2 in Figure 8.9. Let A nd B be the lengths of OP 1 nd OP 2 respectively. The right ngle formed by OP 3 nd P 3 P 1 gives 200 A = sin2, so A = 200 sin Also, 400 B = sin0, so B = 400 sin In tringle OP 1 P 2, the ngle t O is 0 2 = 2. Applying the Lw of Cosines to the tringle OP 1 P 2 gives A 2 +B 2 2ABcos2 = d 2. Sod m.

38 20 Chpter Eight /SOLUTIONS P 2 d P O P 3 Figure yds h which implies tht 38 sin(38 ) = h 100, h = 100sin(38 ) yrds or feet. 33. Let d be the distnce from Hmpton to the point where the bem strikes the shore. Then, tnφ = d 3, so d = 3tnφ miles. Hmpton 3 mi φ Lighthouse d bem Figure () Since B is the point (1,0), the circle hs rdius 1. Thus, sinθ = OE. (b) From the definition of cosθ, we hve cosθ = OA. (c) In ΔODB, we hve OB = 1. Since tn θ = Opp Adj = DB 1, we hve tn θ = DB. (d) LetP be the point of intersection offc ndod. Use the fct thtδofp ndδoep re similr, nd form proportion of the hypotenuse to the one unit side of the lrger tringle. OF 1 = 1 or OF = 1 OE sinθ (e) Use the fct thtδocp4 ndδoap re similr, nd write OC 1 = 1 or OC = 1 OA cosθ (f) Use the fct thtδgoh nd ΔDOB re similr, nd write GH 1 = 1 DB or GH = 1 tnθ

39 STRENGTHEN YOUR UNDERSTANDING If we consider tringle with opposite side of length3nd hypotenuse, we cn use the Pythgoren theorem to find the length of the djcent side s = 4. This gives tringle with sides 3 nd 4, nd hypotenuse. Since tngent is negtive in the fourth qudrnt, tnθ = 3 4. See Figure y 3 θ θ 4 3 Figure From the figure, we see tht so Using clcultor, we find tht = sin 1 (0.83) sin = , = sin 1 (0.83). 37. () Clerly ( 1) 2 +y 2 = 1 is circle with center (1,0). To convert this to polr, use = rcosθ nd y = rsinθ. Then (rcosθ 1) 2 +(rsinθ) 2 = 1 or r 2 cos 2 θ 2rcosθ +1+r 2 sin 2 θ = 1. This mens r 2 (cos 2 θ +sin 2 θ) = 2rcosθ, or r = 2cosθ. (b) 12 o clock (,y) = (1,1) nd (r,θ) = ( 2,π 4), 3 o clock (,y) = (2,0) nd(r,θ) = (2,0), 6 o clock (,y) = (1, 1) nd (r,θ) = ( 2, π 4), 9 o clock (,y) = (0,0) nd(r,θ) = (0, ny ngle ). STRENGTHEN YOUR UNDERSTANDING 1. Flse. Both cute ngles re 4 degrees, ndsin4 = Flse. The ldder is the hypotenuse of right tringle. Using the 60 ngle, we know 3 2 = sin60 = w 16, where w is the height bove ground. Solving this eqution, we get w = Flse. It cn be pplied to ny tringle. 4. True. For emple, to find ngle C, substitute the vlues, b nd c, into c 2 = 2 +b 2 2bcosC. Solve for the ngle C by using lgebr nd thecos 1 function.. True. By the Lw of Cosines, we hve p 2 = n 2 +r 2 2nrcosP, so cosp = (n 2 +r 2 p 2 ) (2nr). 6. True. IfC = 90, thencosc = 0, so by the Lw of Cosines, c 2 = 2 +b 2 2bcosC, becomes c 2 = 2 +b Flse. The Lw of Cosines requires tht the length of two sides be known. 8. True. Use the Lw of Sines nd multiply by the common denomintor. 9. True. Identify the opposite ngles s B ndlnd use the Lw of Sines to obtin LA sinb = BA LA. Thus sinl BA = sinb sinl.

40 22 Chpter Eight /SOLUTIONS 10. True. Use the Lw of Cosines to find the squre of the third side, then tke the squre root to find the length. 11. True. Given two ngles, the third cn be determined since the sum of the ngles in tringle is180. Use the Lw of Sines to crete rtio involving the unknown nd known side lengths. Solve for the unknown side length. 12. Flse. Using the Lw of Sines we find sin60 8 sina = 3 3 > 1. There is no solution fora Flse. The grph of r = 1 is the unit circle. = sina. However, this hs no solution for A becuse it evlutes to Flse. The grph is the ry from the origin forming n ngle of 1 rdin with the positive -is. 1. Flse. The point (3,π) in polr coordintes is ( 3,0) in Crtesin coordintes. 16. Flse. Crtesin coordinte vlues re unique, but polr coordinte vlues re not. For emple, in polr coordintes,(1, π) nd (1,3π) represent the sme point in they-plne. 17. True. The point is on the y is three units down from the origin. Thus r = 3 nd θ = 3π 2. In polr coordintes this is (3,3π 2). 18. Flse. The region is sector of disk with rdius two, centered t the origin. The ngle θ sweeps out n ngle of two rdins, which is bout 1/3 of full circle.