Solutions Key Right Triangles and Trigonometry

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1 CHAPTER 8 Solutions Key Right Triangles and Trigonometry ARE YOU READY? PAGE D. C 3. A 4. E 5. PR RT 10 5 QR ; RS 1 6 PRQ TRS by Vert. Thm. yes; PRQ TRS by SAS 6. AB FE ; BC ED B E by Rt. Thm. yes; ABC FED by SAS 7. x x 16 x x 8 9. x x (3) (x - 1) 1 x x (x - 3) 6 8x x x y y 1 3 4(3) y(1) y 1 m m 9() 18 m (y + 5) -1 y y (z + 4) -z - 4 z x 3 5(3) 8(x) 160 8x x y 4 9 y y(y) 4(9) y 36 y ± a. x ()(8) 16 x 4 c. x (8)(9) 7 x (u)(3) 81 3u u 7 v (3)(3 + u) v (3)(30) 90 v b. x (10)(30) 300 x w (u)(3 + u) w (7)(30) 810 w Let x be height of cliff above eye level. (8 ) 5.5x x 14.5 ft Cliff is about , or 148 ft high. THINK AND DISCUSS 1. Set up the proportion 7 x x, and solve for x. 1 x 7(1) 147 x EXERCISES GUIDED PRACTICE 1. 8 is geometric mean of and 3.. Sketch the 3 rt. with of in corr. positions. 8-1 SIMILARITY IN RIGHT TRIANGLES, PAGES CHECK IT OUT! 1. Sketch the 3 rt. with of in corr. positions. By Thm , RPQ PSQ RSP. By Thm , LJK JMK LMJ. 175 Holt McDougal Geometry

2 3. Sketch the 3 rt. with of in corr. positions. By Thm , BED ECD BCE. 4. Sketch the 3 rt. with of in corr. positions. By Thm , XYZ XWY YWZ. 5. x ()(50) 100 x x ( 1 ) (8) 4 x 9. x (16)(5) 400 x x (10)(6) 60 x z (10)(4) 40 z x (4)(16) 64 x 8 8. x (9)(1) 108 x x (7)(11) 77 x 77 y (6)(4) 4 y x y (0)(0 + 5) 500 x 5 y z (5)(0 + 5) 15 z (6 13 ) (18)(18 + z) z z z 8 y (8)(18 + 8) 08 y RS (64)(60) 3840 RS m x (8)(18) 144 x 1 PRACTICE AND PROBLEM SOLVING 15. By Thm , MPN PQN MQP. 16. By Thm , CAB ADB CDA. 17. By Thm , RSU RTS STU. 18. x (5)(45) 5 x x (5)(8) 40 x 10. x (1.5)(1) 18 x (4 + x) x 18 4x x 3 z 3(4 + 3) 115 z 4 5. x (30)(40) 100 x 0 3 z (70)(40) 800 z x (3)(15) 45 x 3 5 4) 1. x ( 1 x 5 3. x ( (80) 0 3) ( 7 40) 9 0 x _ 3 _ y 4(3) 18 y 8 y (30)(70) 100 y (z)(1.8) z z 7. y (1.8)( ) 56 y 16 x (7.)( ) 144 x 1 7. Let h represent height of tower above eye level. 91 ft 3 in ft (91.5) 5h h 1665 ft Tower is about ft high x x 3 9. ( 5 ) 0 6x x y; x z z y 3. y; x + y v v y 31. x + y; x + y u 33. z; y z z x u x 34. v; v y(x + y) 35. x; u (x + y)x 36. BD (AD)(CD) (1)(8) 96 BD BC (AC)(CD) (16)(5) 80 BC Holt McDougal Geometry

3 38. BD (AC)(CD) ( ) ( ) BD 40. (0.1) (0.03) X 100% 5.5% 39. BC (AC)(CD) 5 CD CD CD _ B is incorrect; proportion should be 1 EF EF a ()(5) 10 a Altitude is about 3. cm long. 43. By Corollary 8-1-3, a x(x + y) and b y(x + y). So a + b x(x + y) + y(x + y). By Distrib. Prop., this expression simplifies to (x + y)(x + y) (x + y ) c. So a + b c. 44a. S W (RS)(ST ) (4)(3) 1 SW ft, or 3 ft 6 in. b. R W (RS)(RT ) (4)(7) 8 RW ft, or 5 ft 3 in. 45. Area of rect. is ab, and area of square is s. It is given that s ab, so s is geometric mean of a and b. 46. Let z be geometric mean of x and y, where x a and y b. So z a b ab, which is a whole number. TEST PREP 47. D XY (8)(11) 88 XY 9.4 ft 48. H BD (9)(4) 36 BD 6 Area 1 (BD)(AC) 1 (6)(13) 39 m 49. A RS (1)(y + 1) y + 1 RS y + 1 CHALLENGE AND EXTEND 50. Let x be length of shorter seg. 8 (x)(4x) 4 x 8 4 x x x 4 Lengths of segs. are 4 in. and 4(4) 16 in. 51. ( 1 ) (x)(x + 5) 84 x + 5x 0 x + 5x (x - 7)(x + 1) x 7 (since x > 0) y (7)(5) 35 y 35 z 5(5 + 7) 60 z Let AD DC a. By Corollary 8-1-3, AB (a)(a) a, and BC (a)(a) a. So AB BC a. Therefore ABC is isosc., so it is a Step 1 Apply Cor in BDE to find BF and BD. E F (BF)(FD) BF BF.14 BD Step Apply Cor in BDE to find BE. B E (BF)(BD) BE Step 3 Apply Cor in BCD to find BC. BD (BE)(BC) BC BC Step 4 Apply Cor in BCD to find CD. C D (BC)(EC) CD Step 5 Apply Cor in ABC to find AC. B C (AC)(CD) AC AC 15.6 cm Step 6 Apply Pyth. Thm. in ABD to find AB. A B B D + A D A B ( ) AB 8.53 cm SPIRAL REVIEW 54. at x-intercept, y 0 3(0) x x at x-intercept, y 0 x + 4 (0) x at x-intercept, y 0 3(0) x x -1 at y-intercept, x 0 3y + 4 6(0) 3y -4 y at y-intercept, x y y at y-intercept, x 0 3y (0) 3y 15 y Holt McDougal Geometry

4 c (3) 6 c () c (7) DEC is a rt., so 30y 90 y 3 m EDC 8(3) DB bisects ADC, so m EDA m EDC AB BC x + 8 4x 8 x x 4 AB (4) TRIGONOMETRIC RATIOS, PAGES CHECK IT OUT! 1a. cos A b. tan B c. sin B a. c. tan cos b. tan 45 s s 1 sin a. DF is the hyp. Given: EF, opp. to given D. Since opp. side and hyp. are involved, use a sine ratio. opp. leg sin D _ EF hyp. DF sin DF DF 17 sin m b. ST is adj. to the given. Given: TU, the hyp. Since adj. side and hyp. are involved, use a cosine ratio. adj. leg cos T ST hyp. TU cos 4 ST (cos 4 ) ST ST 7.06 in. c. BC is adj. to the given. Given: AC, opp. B. Since opp. side and adj. side are involved, use a tangent ratio. opp. leg tan B _ adj. leg AC BC tan 18 1 BC BC ft tan 18 d. JL is opp. the given. Given: KL, the hyp. Since opp. side and hyp. are involved, use a sine ratio. opp. leg sin K _ JL hyp. KL sin 7 _ JL (sin 7 ) JL JL 6.17 cm 5. 1 Understand the Problem Make a sketch. The answer is AC. Make a Plan AC is the hyp. You are given AB, the leg opp. C. Since opp. leg and hyp. are involved, write an equation using a sine ratio. 3 Solve sin C AB AC sin AC AC _ ft sin Look Back Problem asks for AC rounded to nearest hundredth, so round the length to Length AC of ramp is ft. THINK AND DISCUSS 1. Solve sin Solve cos AB AB. 178 Holt McDougal Geometry

5 3. EXERCISES GUIDED PRACTICE 1. sin J LK JL 3. sin C cos A tan C tan N MP MN tan A cos C sin A cos 60 x x tan 30 _ x 3 x tan sin 45 _ s s sin BC is opp. the given. Given: AC, the hyp. Since opp. side and hyp. are involved, use a sine ratio. opp. leg sin A _ BC hyp. AC sin 3 BC 4 4(sin 3 ) BC BC 1.56 in. 19. QR is opp. the given. Given: PQ, adj. to given. Since opp. and adj. sides are involved, use a tangent ratio. opp. leg tan P _ adj. leg QR PQ tan 50 QR (tan 50 ) QR QR 9.65 m 0. KL is adj. to the given. Given: JL, the hyp. Since adj. side and hyp. are involved, use a cosine ratio. adj. leg cos L KL hyp. JL cos 61 KL.5.5(cos 61 ) KL KL 1.1 cm 1. 1 Understand the Problem The answer is XY, opp. the given. Make a Plan You are given WZ, which is twice WY, the leg adj. to W. First, calculate WY. Then, since opp. and adj. legs are involved, write an equation using a tangent ratio. 3 Solve WY 1 WZ 1 (56) 8 ft tan W _ XY WY tan 15 XY 8 XY 8(tan 15 ) ft 4 Look Back Problem asks for XY rounded to nearest inch. Height XY of pediment is 7 ft 6 in. 16. sin cos cos tan PRACTICE AND PROBLEM SOLVING. cos D tan D tan F cos F sin F sin D sin 30 x x 1 tan 60 x 3 x Holt McDougal Geometry

6 30. _ cos 45 s s 31. tan sin cos tan sin cos PQ 11 sin cos 46 _ cm AC AC _ 19. cos in. 39. sin cos 5 33 GH XZ GH 11 sin ft 41. tan KL KL 9.5 tan ft XZ 33 cos in. 4. EF 83.1 tan m 43. sin l l 1.58 sin m 44. If a and b are opp. and adj. leg lengths, tan (m ) a b 1 a b is , so m sin 45 _ s cos 45 s So sine and cosine ratios are. 46. _ 47. cos 30 x 3 sin 60 x cos 30 sine of a h 10 sin ft 49. BC AD 3 tan (90-68) 1. ft sin 30 x x 1 Sine of a 30 is 0.5. _ RS 50. SU cos 49 _ UT cos 49 _ in. cos The tangent ratio is < 1 for measuring < 45 and > 1 for measuring > 45. In a , both legs have same length, so tan If the acute measure increases, opp. leg length also increases, so tangent ratio is > 1. If the acute measure decreases, the opp. leg length also decreases, so tangent ratio is < 1. 5a. AC AB sin C 5 sin ft 7 ft 7 in. 53. From diagram, sin A From diagram, tan D b. AD AB sin ABD 5 sin (90-8).07 ft ft 1 in. 55. Let 1 face be ABC with A the apex; let M be mdpt. of BC. BC BM h tan 5 (48) 753 ft tan 5 56a, b. Check students work. c. 0 d. sin cos tan e. Values in part d should be close to estimate-based values in part b. _ 57a. tan 30 x 3 x 3 3 sin 30 x x 1 cos 30 _ x 3 x 3 1_ So tan 30 _ sin 30 cos 30 b. tan A a, sin A a, cos A b b c c c. sin A cos A a_ c b_ a c c b a b tan A c 58. (sin 45 ) + (cos 45 ) ( ) ( ) 180 Holt McDougal Geometry

7 59. (sin 30 ) + (cos 30 ) ( 1 ) + ( (sin 60 ) + (cos 60 ) ( 3 ) 61a. sin A a, cos A b c c ) + ( ) b. (sin A ) + (cos A ) ( a c) + ( b c) _ a + b c c c 1 c. Derivation of identity uses fact that in a rt., a + b c, which is Pyth. Thm. sin A BC AB ; cos B BC ; sin A cos B; AB A and B are comp.; the sine of an is equal to the cosine of its comp. 6. P + tan 4 + /cos m A 1 ()( tan 4 ) 0.89 m _ cos 51 + _ cos 51 A P cm (7.) ( 7. tan 51 ) cm 64. P sin ft tan 58 A 1 (4) 4 ( 5.00 ft tan 58 ) 65. P sin cos 7.60 in. A 1 (10 sin 7 )(10 cos 7 ) in. 66. sin A BC AB ; cos B BC ; sin A cos B; A and B AB are comp.; sine of an is to cosine of its comp. 67. Tangent of an acute increases as measure of the increases. TEST PREP 68. A 69. H 17 tan ft 70. C cos N NP MN sin M CHALLENGE AND EXTEND 71. AB tan A BC (4x) tan 4 3x + 3 (4 tan 4-3)(x) 3 x 3 4 tan AB 4(5) 0 BC 3(5) AC AC cos A AB (15x) cos 1 5x + 7 (15 cos 1-5)(x) 7 x 7 15 cos AB 5(3) AC 15(3) 45 BC (tan A ) + 1 ( sin A cos A) Int. of a reg. pentagon measure ( 5-5 )(180) 108. In the diagram, m 1 1 (108) 54. Therefore, r _ in. cos csc Y 1 sin Y 1 XZ YZ YZ XZ cot Y 1 tan Y 1 XZ XY XY XZ _ (sin A ) + (cos A ) (cos A ) _ 1 (cos A ) sec Z cos Z 1 XZ YZ YZ XZ SPIRAL REVIEW Possible answers given. 78. (-3, -15), (-1, -9), (0, -6) 79. (-, 11), (0, 10), (, 9) 80. (-, 14), (0, ), (4, ) 81. Trans. Prop. of 8. Reflex. Prop. of 83. Symm. Prop. of Holt McDougal Geometry

8 8-3 SOLVING RIGHT TRIANGLES, PAGES CHECK IT OUT! PAGES _ 1a sin A A is a. ta n -1 (0.75) 37 _ b tan A 14.4 A is 1 b. co s -1 (0.05) 87 _ c. 3. DF DE sin F 14 sin EF DE tan F 14 si n -1 (0.67) 4 tan Acute of a rt. are comp. So, m D Step 1 Find side lengths. Plot pts. R, S, and T. RS 7 ST 7 RT (-3-4 ) + (5 + ) (-7) Step Find measures. m S 90 m R ta n -1 ( 7 7) 45 m T _ 5. 38% A 38% grade means Baldwin St. rises 38 ft for every 100 ft of horiz. dist. m D ta n -1 ( _ 100) 38 1 THINK AND DISCUSS 1. Find RS using Pyth. Thm. Then find m R using m R si n -1 ( 3.5, and find m T using either 4.1) m T co s -1 ( 3.5 or m T 90 - m R. 4.1). co s -1 (0.35) m Z 3. EXERCISES GUIDED PRACTICE sin A A is cos A 10 A is tan A 8 A is ta n -1 (.1) 65 co s -1 ( 5 6) tan A A is cos A 10 A is sin A 10 A is co s -1 ( 1 3) 71 si n -1 (0.5) 30 si n -1 (0.61) 38 ta n -1 (0.09) tan P m P ta n -1 ( 8.9) Acute of a rt. are comp. So m R RP 3.1 sin P ) ) sin ( ta n -1 ( AB 7.4 cos BC 7.4 sin Acute of a rt. are comp. So m C Holt McDougal Geometry

9 15. By Pyth. Thm., YZ tan Y m Y ta n -1 ( ) 38 tan Z m Z ta n -1 ( 8.6) Step 1 Find side lengths. Plot pts. D, E, and F. DE 3 EF 6 DF (- - 4 ) + (- - 1 ) (-6) + (-3) Step Find measures. m E 90 m D ta n -1 ( 6 3) 63 m F Step 1 Find side lengths. Plot pts. R, S, and T. RS 5 ST 6 RT (- - 3 ) + (-3-3 ) (-5) + (-6) Step Find measures. m S 90 m R ta n -1 ( 6 5) 50 m T Step 1 Find side lengths. Plot pts. X, Y, and Z. XZ 7 YZ 7 XY (-3-4 ) + (1 + 6 ) (-7) Step Find measures. m Z 90 m X ta n -1 ( 7 7) 45 m Y Step 1 Find side lengths. Plot pts. A, B, and C. AB BC 4 AC (1 + 1 ) + (5-1 ) Step Find measures. m B 90 m A ta n -1 ( 4 ) 63 m C _ 0. 8% An 8% grade means hill rises 8 m for every 100 m of horiz. dist. m ta n -1 ( _ 100) 8 5 PRACTICE AND PROBLEM SOLVING tan tan 1 A A _ sin _ sin A 1 A _ cos _ cos 1 A A 1 7. si n -1 (0.31) ta n -1 (1) co s -1 (0.8) co s -1 (0.7) ta n -1 (1.55) si n -1 ( 17) Step 1 Find unknown side lengths. JK 3. cos 6.88 LK 3. sin Step Find unknown measure. m L Step 1 Find unknown side length. DF ( 5 ) + ( ) Step Find unknown measures. m D ta n ( -1 5 ) 3 m F ta n ( -1 5 ) Step 1 Find unknown measures. m P co s -1 ( 8.3) m R si n -1 ( 8.3) Step Find unknown side length. QR 8.3 sin P 8.3 sin ( co s -1 ( ) ) Holt McDougal Geometry

10 36. Step 1 Find side lengths. AB is vert., AB 5; BC is horiz., BC 1 By Pyth. Thm., AC Step Find measures. m B 90 m A ta n -1 ( BC AB) si n -1 ( 1 5) 11 m C Step 1 Find side lengths. MN is vert., MN 4; NP is horiz., NP 4 By Pyth. Thm., MP Step Find measures. m N 90 m M ta n -1 ( NP MN) ta n -1 ( 4 4) 45 m P Range of measures is between ta n -1 ( 0) 1 3 and ta n -1 ( 16) ta n -1 (3.5) 74 tan si n -1 ( 3) 4 sin cos cos co s -1 (0.98) sin si n -1 (0.93) cos Assume square has sides of length a. Then either rt. formed by a diag. has legs of length a. So measure of formed by diag. and a side is ta n -1 ( a a) ta n -1 (1) a. Possible answer: m P 40 b. RQ. cm, PQ 3.1 cm c. m P ta n ( -1 RQ PQ) ta n -1 (. 3.1) 35 d. Possible answer: Answer in part c is likely more accurate, since it is easier to measure lengths to the nearest tenth than to measure to the nearest degree. 47a. m 1 ta n -1 ( _ 100) 8 5 b. m c. h ft, or 31 ft 1 in. 48. si n -1 ( 3 sin ( 90 - ta n -1 ( 8 5) 37, si n -1 ( 4 100) ) 5) si n -1 ( 13) 5 3, si n -1 ( 1 13) si n -1 ( 17) 8 8, si n -1 ( 15 17) ta n -1 ( 45 8) 58, ta n -1 ( 8) Acute measure changes from about 58 to about 73, an increase by a factor of 1.6. _ 5. m ta n -1 ( 100) a. AB (6 + 1 ) + (1-0 ) BC (0-6 ) + (3-1 ) AC (0 + 1 ) + (3-0 ) b. A C + B C A B So ABC is a rt., and C is the rt.. c. m A si n -1 ( BC AB) si n ( ) _ si n ( ) 63 m B 90 - m A m BDC ta n -1 ( 7) m STV ta n -1 ( ) m DGF m DGH si n -1 (.4 4.4) m LKN ta n -1 ( 4.8) tan 70 > tan 60 ; possible answer: consider rt., 1 with a 60 and 1 with a 70. Suppose that legs adj. to these have length 1 unit. Leg opp. 70 will be longer than leg opp. 60. So tan 70 is greaater than tan ta n -1 (m) ta n -1 (3) ta n -1 (m) ta n -1 ( 3) y 4x + 3 y 4 5 x ta n -1 (m) ta n -1 ( 4 5) Since is not a rt.., trig. ratios do not apply. 63. No; possible answer: you only need to know side lengths. You can use Pyth. Thm. to find 3rd side length or use trig. ratios to find acute measures. 184 Holt McDougal Geometry

11 64. AD AC cos A 10 cos ( ta n -1 ( 6 BD BC cos B 10) ) cos ( ta n -1 ( 10 6 )) 3.09 CD BC sin B 6 sin ( ta n -1 ( 10 6 )) 5.14 (8.57)(3.09) 6 (5.14) (8.57)( ) 100 (10) (3.09)( ) 36 (6) TEST PREP 65. D 66. J 67. A ta n -1 ( ) 7 ta n -1 ( 0) 3 9 CHALLENGE AND EXTEND 69. LH 10 sin J 0 sin 5 sin J sin 5 m J si n -1 ( sin 5 ) BD 3. tan A 8 cos 64 tan A.5 cos 64 m A ta n -1 (.5 cos 64 ) Let A be an acute with m A co s -1 (cos 34 ). Then cos A cos 34. Since cos is a 1-to-1 function on acute measures, m A Since tan is a 1-to-1 function on acute measures, x tan [ta n -1 (1.5)] x Since sin is a 1-to-1 function on acute measures, y sin ( si n -1 x ) y x _ 74. y 40 sin ( ta n -1 ( 6 100) ).40 ft 75. Possible answer: The expression sin -1 (1.5) represents an measure that has a sine of 1.5. The sine of an acute of a rt. must be between 0 and 1, so the expression sin -1 (1.5) is undefined. 76. Let BD be altitude. Then Area 1 (base)(height) 1 (AC)(BD) 1 (b)(c sin A) 1 bc sin A. SPIRAL REVIEW 77. false; true; 3.5 in False; rainfall decreased from April to May. 80. B E 81. AB 37 x + 11 DE AC DF 6 x 3y y x 13 3y + 7 y + 6 y DF sin cos tan READY TO GO ON? PAGE x (5)(1) x ( 5 ) ( 15 8 ) _ (1 5 ) (4)(x + 4) 70 4x x x 6 y 4x 4(6) 144 y 1 z (4 + x)(x) (30)(6) 180 z x (.75)(44) x (4)(8) 3 x 3 4 y (4)(1) 48 y z (8)(1) 96 z x 36 1x x 3 y 1(1 + x) (1)(15) 180 y z (1 + x)x (15)(3) 45 z (AB) (BC)(BD) ()(30) 660 AB m 8. Let legs of have length x. tan 45 x x 1 9. Let have side lengths x, x 3, x. sin 30 x x cos 30 _ x 3 x sin cos tan QR in. tan AB 6 cos m 16. LM 4. sin cm 17. m A BC tan AC sin Holt McDougal Geometry

12 18. HJ m H ta n -1 ( _ ) 56 m J ta n -1 ( _ 10.5) m Z XY 5.1 cos YZ 5.1 sin ta n -1 ( 18) ANGLES OF ELEVATION AND DEPRESSION, PAGES CHECK IT OUT! 1a. 5 is formed by a horiz. line and a line of sight to a pt. below the line. It is an of depression. b. 6 is formed by a horiz. line and a line of sight to a pt. above the line. It is an of elevation.. Let A represent airport and P represent plane. Let x be horiz. distance between plane and airport. tan x x 3500 tan ft 3. Let T represent top of tower and F represent fire. Let x be horiz. distance between tower and fire. By Alt. Int. Thm., m F 3. tan 3 90 x x ft tan 3 4. Step 1 Let P represent plane, and A and B represent two airports. Let x be distance between airports. Step Find y. By Alt. Int. Thm., m CAP 78. In APC, tan 78 1,000 y y 1, ft tan 78 Step 3 Find z. By Alt. Int. Thm., m CBP 19. In BPC, tan 19 1,000 z z 1,000 34,850.6 ft tan 19 Step 4 Find x. x z - y 34, ,300 ft THINK AND DISCUSS 1. It increases, because height of skyscraper is constant and horiz. dist. is decreasing.. EXERCISES GUIDED PRACTICE 1. elevation. depression 3. 1 is formed by a horiz. line and a line of sight to a pt. above the line. It is an of elevation. 4. is formed by a horiz. line and a line of sight to a pt. below the line. It is an of depression is formed by a horiz. line and a line of sight to a pt. above the line. It is an of elevation is formed by a horiz. line and a line of sight to a pt. below the line. It is an of depression. 7. Let h be height of flagpole. tan 37 _ h 4. h 4 tan ft 186 Holt McDougal Geometry

13 8. Let H represent helicopter and A represent accident. Let x be horiz. dist. between helicopter and accident. By Alt. Int. Thm., m A 18. tan x x ft tan Step 1 Let T represent top of canyon, and A and B represent near and far sides of river. Let w be width of river. Step Find y. By Alt. Int. Thm., m CAT 74. In ATC, tan 74 _ 191 y y 191 tan m Step 3 Find z. By Alt. Int. Thm., m CBT 58. In BTC, tan 58 _ 191 z z 191 tan m Step 4 Find w. w z - y m PRACTICE AND PROBLEM SOLVING 10. 1: of depression 11. : of elevation 1. 3: of elevation 13. 4: of depression 14. h tan m 15. x _ ft 16. z y - x tan tan 74-1 tan mi 17. true 18. true 19. false; of elevation gets closer to true 1. 1 and 3. m 30 (given) m m (comp. ) m 3 m 1 60, m 4 m 30 (Alt. Int. Thm.) 3. Possible answer: As a hot air ballon descends vertically, of depression to an object on the ground decreases. 4. By Alt. Int. Thm., of depression ta n -1 ( _ ) 73 5a. x ft b. z y - x tan tan tan ft 6. When the of elevation is exactly 45, the length of the shadow will be the same as the length of telephone pole, since a rt. isosc. is formed and tan a. x ft b. v s tan 31 t TEST PREP 8. D dist ft tan 35 t s v s 9. J height 93 tan ft 30. The of elevation increases as Jim moves closer to trail marker. CHALLENGE AND EXTEND 31. Let x and y be dists. from Jorge and from Susan to foot of Big Ben; let h be height of Big Ben. Jorge: h x tan 49.5 Susan: h y tan 65 y x - 38 h x tan 49.5 (x - 38) tan tan 65 x(tan 65 - tan 49.5 ) x 38 tan 65 (tan 65 - tan 49.5 ) h x tan tan 65 (tan 49.5 ) 98 m (tan 65 - tan 49.5 ) 3. Speed 500 mi h 1 h 580 ft 44,000 ft/min 60 min 1 mi Let time until over lake be t. Then horiz. dist to lake is s 44,000t 14,000 tan 6 14,000 t 44,000 tan 6 3 min. 33. h x tan 5 (10 - x) tan x(tan 5 + tan ) 10 tan x 10 tan tan 5 + tan h x tan 5 10 tan (tan 5 ) tan 5 + tan mi 1318 ft 34. h y - x 46 tan 4-46 tan ft or 6 ft 6 in. 187 Holt McDougal Geometry

14 SPIRAL REVIEW 35. Let x and y be dists. run by Emma and mother in time t. When they meet, x + y 1 6t + 4t 1 10t 1 t 0.1 h or 6 min 36. Let p be original price. discounted price 0.7p price after coupon 0.85(0.7p) p (0.7) 30 Original price was $ rhombus, square 38. rectangle, square 39. rectangle, rhombus, square 41. x x x 16 x x 3z 4 3z z rectangle, rhombus, square 4. y x 5 3 y 5x 3 _ 5(4) LAW OF SINES AND LAW OF COSINES, PAGES CHECK IT OUT! 1a. b. c. tan cos sin a. _ sin N MP sin M NP sin 39 sin 88 NP NP sin 39 sin 88 NP sin 88 sin c. _ sin X YZ _ sin Y XZ _ sin X sin sin X 4.3 sin m X si n -1 ( sin ) 3a. D E D F + E F - (DF)(EF) cos F (16)(18) cos 1 D E DE 6.5 b. J L J K + K L - (JK)(KL) cos K (15)(10) cos K cos K cos K cos K _ m K co s -1 ( _ 300) c. Y Z X Y + X Z - (XY)(XZ) cos X (10)(4) cos 34 Y Z YZ 7.0 b. _ sin L JK _ sin K JL _ sin L _ sin sin L 6 sin m L si n -1 6 sin 15 ( 10 ) 9 d. m A _ sin A BC _ sin B AC sin 69 sin AC AC sin sin 67 AC 18 sin 67 sin d. P Q Q R + P R - (QR)(PR) cos R (10.5)(5.9) cos R cos R cos R cos R m R co s -1 ( 13.9) Step 1 Find length of cable. A C A B + B C - (AB)(BC) cos B (31)(56) cos 100 A C AC 68.6 m Step Find angle measure between cable and ground. _ sin A BC _ sin B AC _ sin A _ sin sin A 56 sin m A si n sin 100 ( ) Holt McDougal Geometry

15 THINK AND DISCUSS 1. m A.. 6. sin sin EXERCISES GUIDED PRACTICE 1. tan sin cos tan cos sin R ST sin S RT sin 36 sin RT RT sin sin 70 RT 15 sin 70 sin _ sin F DE _ sin D EF _ sin F sin sin F 0 sin F si n -1 0 sin 84 ( 31 ) _ sin B _ AC sin C AB _ sin B _ sin sin B 14 sin B si n sin 101 ( 0 ) 43 tan P R P Q + Q R - (PQ)(QR) cos Q (6)(10) cos Q cos Q cos Q cos Q _ m Q co s -1 ( _ 10) cos M N L M + L N - (LM)(LN) cos L (30)(5) cos77 M N MN Holt McDougal Geometry

16 15. A B A C + B C - (AC)(BC) cos C (8)(11) cos 131 A B AB Think: Find each using Law of Cosines (4)(30) cos cos cos 1 m 1 co s -1 ( ) (0)(30) cos cos cos m co s -1 ( 100) (4)(0) cos cos cos 3 m 3 co s -1 ( _ 960) PRACTICE AND PROBLEM SOLVING 17. cos tan tan sin sin cos tan cos sin _ sin C _ sin m C si n sin 1 ( _ 10. ) sin 17 _ sin _ sin 140 sin PR PR _ 9 JL 8.5 sin 135 JL _ 9 sin 0 sin 17 sin sin 56 sin _ sin J 11.7 EF EF sin sin 47 m J s in -1 sin 56 ( 61 sin ) _ sin X 3.6 sin m X si n sin 78 ( 3.9 ) A B (13)(5.8) cos AB (14.7)(6.8) cos Z cos Z cos Z m Z co s -1 ( ) (1)(13) cos R cos R cos R m R co s -1 ( _ 88 31) E F (8.4)(10.6) cos EF L M (10.1)(1.9) cos LM (13)(14) cos G cos G cos G m G co s -1 ( _ 364) A B (108)(55) cos AB 9.6 _ sin B sin m B si n sin 59 ( ) _ sin B _ sin A b a sin sin a a 3. sin cm sin 40. c a + b - ab cos C (9.5)(7.1) cos c 1.8 in. 41. b a + c - ac cos B (.)(4) cos B cos B cos B m B co s -1 ( ) _ sin C _ sin A c a _ sin C sin m C si n sin 45 ( 10.3 ) No; 3 measures do not uniquely determine a. There is not enough information to use either Law of Sines or Law of Cosines. 44. c a + b - ab cos C a + b - ab cos 90 a + b Law of Cosines simplifies to Pyth. Thm. 45. Let of turn be 1 and let be opp. 6-km side (3)(4) cos cos 11-4 cos m co s -1 ( ) m m co s -1 ( ) Holt McDougal Geometry

17 46. Step 1 Find 3rd side length. Think: Use Law of Cosines. x (5)(9) cos x 11.6 cm Step Find perimeter. P cm 47. Step 1 Find nd side length. Think: Use Sum Thm., Law of Sines. m (93 + 4) 63 sin 63 sin 4 16 x x 16 sin ft sin 63 Step Find 3rd side length. Think: Use Law of Sines. sin 63 sin y y 16 sin ft sin 63 Step 3 Find perimeter. P ft 48. Step 1 Find nd side length. Think: Use Law of Sines. sin 45 sin _ x x 7.3 sin in. sin 45 Step Find 3rd side length. Think: Use Sum Thm., Law of Sines. m ( ) 0 sin 45 sin y y 7.3 sin in. sin 45 Step 3 Find perimeter. P Figure shows two possible positions for C. Since BC BC', C BC'C, so C and BC'A are supp. _ sin C sin m C si n -1 1sin 30 ( 9 ) 4 m BC'A m C a. Think: Use Sum Thm. m F m F m F 91 b. sin 91 sin AF AF 18.3sin 38 sin mi sin sin 51 BF BF 18.3sin 51 sin mi c. Dist. speed time AF 150 t 1 BF 150 t BF - AF 150( t 1 - t ) t 1 - t BF - AF _ h or 1. min 51. Given is opp. one of given sides, so use Law of Sines. 5. Given is included by given sides, so use Law of Cosines. 53. One of given is opp. given side, so use Law of Sines. 54a. RS ST RT b. R, because it is opp. the longest side. c. S T R S + R T - (RS)(RT) cos R ( 13 )(5) cos R (cos R) R co s -1 ( ) B C (6.46)(7.14) cos BC cm A B (3.86)(7.14) cos AB cm _ sin ABE _ sin m ABE si n sin 138 ( ) _ sin EBC _ sin m EBC si n sin 104 ( ) m ABC m ABE + m EBC A is incorrect; possible answer: the fraction on the right side of the proportion is incorrect. It should be sin 70 sin 85, as in B. 1 x 57a. y + h b. b c. a c - cx + x + h d. a c + b - cx e. b cos A f. Subst. 58. No; possible answer: to use Law of Sines, you need to know at least 1 side length and measure opp. that side. 191 Holt McDougal Geometry

18 TEST PREP 59. A A B (1)(14) cos AB 5.7 cm Nearest given value is 5.5 cm. 60. H 61. C m Y ( ) 0 sin 0 sin XY XY 100 sin 5 14 m sin 0 CHALLENGE AND EXTEND 6. A B A C + B C - (AC)(BC) cos ACB ( + 3 ) ( + 4 ) + (3 + 4 ) - ( + 4)(3 + 4) cos ACB cos ACB cos ACB m ACB co s -1 ( 84) Let pts. be A(-1, 1), B(1, 3), and C(3, ) AB + 8 ; AC ; BC B C A B + A C - (AB)(AC) cos A ( 8 )( 17 ) cos A (cos A) m A co s ( ) Let P be position of boat after 45 min 0.75 h. Given information: AB 5 mi, AP (6 mi/h)(0.75 h) 4.5 mi, A B P A B + A P - (AB)(AP) cos A (5)(4.5) cos BP 9.1 mi SPIRAL REVIEW 65. 3n 66. n n Alt. Ext. Thm. 69. Alt. Int. Thm. 70. Same-Side Int. Thm. 71. Alt. Ext. Thm VECTORS, PAGES CHECK IT OUT! 1a. Horiz. change along u is -3 units. Vert. change along u is -4 units. So component form of u is -3, -4. b. Horiz. change from L to M is 7 units. Vert. change from L to M is 1 unit. So component form of LM is 7, 1.. Step 1 Draw vector on a coord. plane. Use origin as initial pt. Then (-3, 1) is terminal pt. Step Find magnitude. Use Dist. Formula. -3, 1 (-3-0 ) + (1-0 ) Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find direction. Draw rt. ABC as shown. A is formed by vector and x-axis, and tan A 3 7. So m A ta n -1 ( 3 7) 3 4a. PQ RS (same magnitude and direction) b. PQ RS and XY MN (same or opp. direction) 5. Step 1 Sketch vectors for kayaker and current. Step Write vector for kayaker in component form. It has magn. 4 mi/h and makes of 70 with x-axis. cos 70 x, so x 4 cos sin 70 y, so y 4 sin Kayaker s vector is 1.37, Step 3 Write vector for current in component form: 1, 0. Step 4 Find and sketch resultant vector AB. Add components of kayaker s vector and current s vector. 1.37, , 0.37, 3.76 Step 5 Find magn. and direction of resultant vector. Magn. of resultant vector is kayak s actual speed..37, 3.76 (.37-0 ) + ( ) 4.4 mi/h measure formed by resultant vector gives kayak s actual direction. tan A _ , so m A ta n -1 ( _.37) , or N 3 E. 19 Holt McDougal Geometry

19 THINK AND DISCUSS 1. It does not have a direction.. Pyth. Thm. 3. Possible answer: Write each vector in component form and add horizontal and vertical components. 4. EXERCISES GUIDED PRACTICE 1. equal. parallel 3. magnitude 4. Horiz. change from A to C is 5 units. Vert. change from A to C is 3 units. So component form of AC is 5, Horiz. change from M to N is 8 units. Vert. change from M to N is -8 units. So component form of MN is 8, Horiz. change from P to Q is units. Vert. change from P to Q is 5 units. So component form of PQ is, Step 1 Draw vector on a coord. plane. Use origin as initial pt. Then (1, 4) is terminal pt. Step Find magnitude. Use Dist. Formula. 1, 4 (1-0 ) + (4-0 ) Step 1 Draw vector on a coord. plane. Use origin as initial pt. Then (-3, -) is terminal pt. Step Find magnitude. Use Dist. Formula. -3, - (-3-0 ) + (- - 0 ) Step 1 Draw vector on a coord. plane. Use origin as initial pt. Then (5, -3) is terminal pt. Step Find magnitude. Use Dist. Formula. 5, -3 (5-0 ) + (-3-0 ) Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find direction. Draw rt. ABC as shown. A is formed by vector and x-axis, and tan A 6 4. So m A ta n -1 ( 6 4) Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find direction. Draw rt. ABC as shown. A is formed by vector and x-axis, and tan A 1 5. So m A ta n -1 ( 1 5) Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find direction. Draw rt. ABC as shown. A is formed by vector and x-axis, and tan A 3 6. So m A ta n -1 ( 3 6) CD EF (same magn. and direction) 14. CD EF and AB GH (same or opp. direction) 15. RS XY (same magn. and direction) 16. RS XY and MN PQ (same or opp. direction) 17. Step 1 Sketch vectors for stages of hike. Step Write vector for 1st stage in component form. It has magn. mi and makes of 50 with x-axis. cos 50 x, so x cos sin 50 y, so y sin Vector for 1st stage is 1.9, Step 3 Write vector for nd stage in component form: 3, 0. Step 4 Find and sketch resultant vector AB. Add components of 1st- and nd-stage vectors. 1.9, , 0 4.9, 1.53 Step 5 Find magn. and direction of resultant vector. Magn. of resultant vector is straight-line dist. to campsite. 4.9, 1.53 (4.9-0 ) + ( ) 4.6 mi measure formed by resultant vector gives direction of hike. tan A _ , so m A ta n -1 ( _ 4.9) , or N 70 E. 193 Holt McDougal Geometry

20 PRACTICE AND PROBLEM SOLVING 18. 9, , , Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find magnitude. Since vector is horiz., -, Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find magnitude. Use Pyth. Thm. 1.5, Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find magnitude. Use Pyth. Thm..5, Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find direction. m A ta n -1 ( ) 1 5. Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find direction. m A ta n -1 ( 3.5) Step 1 Draw vector on a coord. plane. Use origin as initial pt. Step Find direction. m A ta n -1 ( 5 ) DE LM 9. RS UV 30. RS UV AB and CD XY 8. All 4 vectors are. 31. Step 1 Write airplane s vector in component form. x 00 cos ; y 00 sin Airplane s vector is 84.54, Step Write windspeed vector in component form. x 40 cos ; y -40 sin Windspeed vector is 8.84, Step 3 Find resultant vector AB. Add components of airplane s and windspeed vectors , , , Step 4 Find magn. and direction of resultant vector , km/h m A ta n -1 ( , or N 36 E ) 3. 1, + 0, , + 6 1, , 4 + 5, , 4 + (-), 34. 0, 1 + 7, , , , 3 + -, (-), 3 + (-1) 6, 36. Yes; possible answer: if you use head-to-tail method in both orders, you end up with a and its diag. Resultant vector is the diag. See figures below. 37a. Let FG be a vert. line. Use Alt. Int Thm. m F m GFH + m GFX b. H X (50)(41) cos HX 68.9 mi/h c. _ sin FHX sin m FHX si n sin 98 ( 68.9 ) 36 d. direction E of N, or N 81 E cos 4, 15 sin , cos 9, 7. sin 9 7.1, direction relative to x-axis cos 33, 1.1 sin , direction relative to x-axis cos 68, 5.8 sin 68., 5.4 4a. 10 sin lb b. 10 sin lb c. Taneka; she applies more vert. force. 43a. Prob. of 1 on 1st draw is 1 ; prob. of then drawing 4 is 1. So Prob. ( 1, ) b. Prob. (vector to 1, ) Prob. ( 1, or, 4 ) Holt McDougal Geometry

21 44a. a - b 4, -; -, -1, u 4 4 direction of u w direction of w ta n -1 ( 3 ) 56 b. a 4, - - b -, -1 a +(- b ) is shown below 46. v 3 3 direction of v z direction of z ta n -1 ( 1 4) Pass pattern vectors are 0, 10 and 10, 0. Resultant vector is 0, , 0 10, 10. Magn. of resultant is ; Direction of resultant is ta n -1 ( 10) 45. Jason s move is equivalent to resultant Possible answers given. 50. Think: Change sign of one component only. 3, 6 has same magn. but different direction. Think: Multiply both components by the same factor. -6, 1 has same direction but different magn , -5 has same magn. but different (opp.) direction. 4, 10 has same direction but different magn , 11 has same magn. but different (opp.) direction. 4, -5.5 has same direction but different magn. 53. u + v 1 +.5, + (-1) 3.5, 1 u + v direction of u v ta n -1 ( 3.5) u + v , 7 + (-3.1).8, 3.9 u + v direction of u v ta n -1 ( 3.9.8) u + v 6 + (-), , 4 u + v direction of u + 4) 45 v , 8 + (-.1) 4, u + v ta n -1 ( 4 u + v direction of u + 57a. v 1, 3 ; v, 6 a+(-b) v ta n -1 ( ) 56 a -b b. v ; v v is twice the magnitude of v. 1) 7 c. direction of v ta n -1 ( 3 v ta n -1 ( 6 direction of Directions are the same. ) ta n -1 ( 3 d. Multiply each component by k. e. - v (-1) v (-1) x, y (-1)x, (-1)y -x, -y 1) If u > v, resultant points due west, with magn. u - v. If v > u, resultant points due east, with magn. v - u. If u v, resultant is 0, A line seg. has magnitude (or length), but no direction. A ray is a part of a line that continues indefinitely in one direction. Thus it has direction and infinite magnitude. A vector has both direction and magnitude. TEST PREP 60. C w (-), 1 6. C G ta n -1 ( 9 7) AB ( ) + (- - 6 ) CHALLENGE AND EXTEND 64. -, 3 is in nd quadrant, so direction is between 90 and 180. direction ta n -1 ( -) , 0 lies along negative x-axis, so direction , -3 is in 3rd quadrant, so direction is between 180 and 70. direction ta n -1 ( -5) Let v x, y be required velocity vector. Then x, y + 4, 0 10 cos 0, 10 sin 0 x cos 0 x 10 cos mi/h y 10 sin mi/h v mi/h bearing ta n -1 ( _ 5.40) 3.4 3, or N 58 E 68. v x, y 3 cos 60, 3 cos , cos 40, 4 sin 40 x 3 cos cos km y 3 sin sin km v km bearing ta n -1 ( 10.56) , or N 64 E 195 Holt McDougal Geometry

22 SPIRAL REVIEW 69. x - y -5 y 3x + 1 y x + 5 y 3x + 1 Lines intersect at (, 7). 70. x - y 0 y + x 8 y 0.5x y -0.5x + 4 Lines intersect at (4, ). 71. x + y 5 3y + 15 x y -x + 5 y 3 x - 5 Lines intersect at (6, -1). 7. NP 3JL 73. Area (3) (6) Perim. 3(1) 36 cm 54 c m 74. B C (3.5)(4) cos BC _ sin B sin m B si n -1 4 sin 50 ( _ 3.0 ) m C ( ) 57 READY TO GO ON? PAGE By Alt. Int. Thm., dist ft. tan 34. height 6 tan m 3. _ sin A sin _ A si n sin 118 ( 0 ) sin 41 sin 84 7 GH GH _ 7 sin 84 sin m X (9 + 6) 6 sin 6 sin 9 8 XZ XZ _ 8 sin 9 sin U V (1)(9) cos UV (5)(6) cos F cos F cos F m F co s -1 ( 60) Q S (10.5)(6) cos QS , , , direction ta n -1 ( 1 ) direction ta n -1 ( 3 5) direction ta n -1 ( 4 4) Let x, y be resultant vector. x, y 6 sin 3, 6 cos 3 + 8, 0 x 6 sin km y 6 cos km dist km direction ta n -1 ( 11.18) , or N 66 E 196 Holt McDougal Geometry

23 STUDY GUIDE: REVIEW, PAGES VOCABULARY 1. component form. equal vectors 3. geometric mean 4. angle of elevation 5. trigonometric ratio LESSON PRQ RSQ PSR 4) 7. x ( 1 (100) 5 x x (5)(7) 35 x 35 y 5(5 + 7) 60 y z 7(5 + 7) 84 z ( 6 ) (1)(1 + x) x x 5 z (5)(1 + 5) 30 z 30 LESSON 8-1. sin UV UV 11 sin m 14. cos 33 _ XY 1.3 XY 1.3 cos cm LESSON m C AB 5. cos 4.8 AC 5. sin m H ta n -1 ( 3.5) m G HG m S RS 3.5 sin RT 3.5 tan m Q ta n -1 ( ) 41 m N QN x (3)(17) 51 x (x)(1) 36 1x x 3 y (3)(3 + 1) 45 y z (1)(3 + 1) 180 z y (1)(5) 5 y cos 9 PR 7. PR 7. cos m 15. tan JL JL 1.4 tan cm LESSON of depression 1. of elevation. height 5.1 tan 8 36 ft 3. horiz. dist. 3 tan m LESSON 8-5 _ 4. sin Z sin m Z si n -1 4 sin 40 ( _ 7 ) sin E F (14)(1) cos EF (6)(1) cos Q cos Q cos Q m Q co s -1 ( _ 144) LESSON AB - - 5, , 9. MN -1 - (-), , RS -, , , , direction ta n -1 ( 5 4) direction ta n -1 ( 7) 16 _ sin 130 MN MN 16 sin 130 sin Holt McDougal Geometry

24 36. plane s vector 600 cos 35, 600 sin 35 crosswind vector 50, 0 resultant vector 600 cos , 600 sin , speed mi/h direction ta n -1 ( , or N 58 E ) CHAPTER TEST, PAGE (x)(8) 16 8x x z ( + 8) 0 z 0 5. x (6)(1) 7 x 7 6 z 6(6 + 1) 108 z ( 30 ) 10(10 + x) x 0 10x x z ( + 10) 4 z 4 6 y 8( + 8) 80 y y 1(6 + 1) 16 y y ()(10) 0 y Let have 5. Let have sides x, x 3, x. cos 60 x x 1 sides s, s, s. sin 45 _ s s _ 6. tan 60 x 3 x 8. AB 9 cos cm 10. m ta n -1 ( ) 19 _ 1. sin B sin m B si n -1 4 sin 85 ( _ 9 ) PR 4.5 sin m 9. FG 6.1 tan in. 11. horiz. dist. 910 tan ft 13. sin 35 _ sin RS (10)(15) cos M cos M cos M m M co s -1 ( _ ) 3 RS 11 sin 108 sin , , , direction ta n -1 ( 5 3) direction ta n -1 ( 1 4) boat s vector 3.5 cos 50, 3.5 sin 50 current s vector, 0 resultant vector 3.5 cos 50 +, 3.5 sin ,.68 speed mi/h direction ta n -1 ( _.68 3, or N 58 E 4.5) 198 Holt McDougal Geometry