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Practice Test Chapter 5 For Credit show all work Know the following: Trigonometric Ratios Sine (sin = ratio) Cosine (cos = ratio) = opposite side /hypotenuse = adjacent side / hypotenuse = y/r = x/r = 1/csc = 1/ sec Inverse: sin-1(ratio) = angle Inverse: cos-1(ratio) = angle Cosecant (csc ) Secant (sec ) = r/y = r/x = 1/ csc = 1/ cos Values Quadrant I sin cos tan csc sec cot 30 0 1 2 3 2 3 3 2 2 3 3 45 0 2 2 2 2 1 2 2 1 2 3 3 2 3 3 60 0 1 3 3 2 2 Formulas Pythagorean Theorem Use: for all right triangles. c2 a2 b2 Area: A 1 bh 2 Page 1 of 11 Tangent (tan = ratio) = opposite side/ adjacent side = y/x = 1/ cot Inverse: tan-1 (ratio) = angle Cotangent (cot ) = x/y = 1/ tan 3 Law of Sines Use: Angle (must be less than 90o) and opposite side, and another side or angle; a b c sin A sin B sin C 1 bc sin A 2 1 Area: K ac sin B 2 1 K ab sin C 2 K Law of Cosines Use: 3 sides or Angle and 2 non-opposing sides. c 2 a 2 b 2 2ab cos C b 2 a 2 c 2 2ac cos B a 2 b 2 c 2 2bc cos A 1 sin B sin C K a2 2 sin A 1 sin A sin C Area: K b 2 2 sin B 1 sin A sin B K c2 2 sin C Practice Test Chapter 5 For Credit show all work Law of Sines Ambiguous Case (A copy of the Ambiguous Case will be supplied during the test) Case I: A <900Case I: A <900 a<b a≥b No solution One Solution Two Solutions One solution Case II: A ≥900 a≤b a>b No solution One Solution Use the figure on the right for questions 1 3. a< b sin a a = b sin a a > b sin a B 9 a C 4 1. Find the value of the sine for A 2. Find the value of the cotangent of A 3. Find the value of the secant of A 4. If the tan = -3 find the cot 5. Find the tan 1800 6. Find the exact value of cos 3300 For exercises 7 and 8, refer to River to waterfall the figure on the right. The angle of elevation from Height of the the far side of the pool to the 0 waterfall top of the waterfall is 68 and the distance is 200 feet. A 1. ___________ 2. __________ 3. __________ 4. _________ 5. _________ 6. _______________ pool of water 7. Find the height of the waterfall to the nearest foot. 8. Find the width across the pool to the nearest foot. 3 9. If 00≥ x ≤ 3600, solve sin x 2 Page 2 of 11 7. _____ 8. __________ 9. ___________ Practice Test Chapter 5 For Credit show all work 10. Given the triangle on the right, find B to the nearest tenth of a degree if b = 12, and c = 18. 10. ________ B c a C b A For exercise 11 and 12, round the numbers to the nearest tenth. 11. If ∆ABC, A= 470, B=580, b=23. Find a 11. ______________ 0 12. If C=37.2 , a =17.9 and b = 22.3, find the area of ∆ABC 12. _______________ 13. Determine the number of possible solutions if A=470, a = 2 13. ____________ and b=4. For exercise 14 and 15, round the numbers to the nearest tenth. 14. In ∆ABC, A=670, b = 10 and c= 5, Find a 14. _________ 15. In ∆ABC, a=8, b = 6 and c= 12, Find C 15. __________ 16. If a=18, b = 22 and c =30, find the area of ∆ABC 16. _________ 17. Aviation A traffic helicopter is flying 1000 feet above the 17. __________ downtown area. To the right, the pilot sees the baseball stadium at an angle of depression of 630. To the left the pilot sees the football stadium at an angle of depression of 180. Find the distance between the two stadiums. 18. Railways: The steepest railway in the world is the 18 A) __________ Katoomba Scenic Railway in Australia. The passenger car is 18 B) __________ pulled up the mountain by twin steel cables. It travels along the track 1020 feet to obtain a change of altitude of 647 feet. a) Find the angle of elevation b) How far does the train travel horizontally. 19. Engineering The escalator of St. Petersburg Metro in 19. ________ Russia has a vertical rise of 195.8 feet. If the angle of elevation of the escalator is 100 21’ 36”, find the length of the escalator? 20. Architecture: A flagpole 40 feet high stands on top of the 20. _________ Wentworth Building. From the point in front of Bailey’s Drugstore, the angle of elevation for the top of the pole is 540 54’ and the angle to the bottom of the pole is 470 30’. How high is the building? Page 3 of 11 Practice Test Chapter 5 For Credit show all work Solutions Use the figure on the right for questions 1 3. B 9 a C 4 A 1. Find the value of the sine for A Sin A = opposite side / hypotenuse = a/c. Hypotenuse = 9 Find opposite side a a2 b2 c2 1. ___________ 65 9 a 2 42 92 a 2 9 2 4 2 81 16 65 a 2 75 a 2 65 a 65 Find sin A a 65 sin A 9 9 2. Find the value of the cotangent of A Cot A = adjacent side / opposite side = b/a From the prior problem we know that a 65 B =4 4 Cot A = which is improper ¾ credit for this answer. 65 Complete answer 4 65 4 65 65 65 65 3. Find the value of the secant of A Sec A = hypotenuse / adjacent side = c/b {inverse of the cos} Adjacent = 4 Hypotenuse = 9 Sec A = 9/4 or 2.25 Page 4 of 11 2. __________ 4 65 65 3. __________ 9/4 or 2.25 Practice Test Chapter 5 For Credit show all work 4. If the tan = -3 find the cot Cot = 1/ tan tan = -3 Cot = 1/(-3) = -1/3 5. Find the tan 1800 Tan = y/x, use calculator or coordinates (-1,0) Tan 180 = 0/(-1)=0 6. Find the exact value of cos 3300 Since the question asks for exact value use the radical 3300 is in Quadrant IV 3300 has a reference angle in Quadrant I 300 Cosine = x/r and x is positive in both Quadrant I and IV 3 Cos 3300 = Cos 300 = 2 For exercises 7 and 8, refer to River to waterfall the figure on the right. The angle of elevation from Height of the the far side of the pool to the waterfall top of the waterfall is 680 and the distance is 200 feet. 4. _________ -1/3 5. _________ 0 6. _______________ 3 2 pool of water 7. Find the height of the waterfall to the nearest foot. x 200 feet 680 The angle of elevation is 680 The height is the opposite side =x The hypotenuse is 200 feet x sin 680 200 200 sin 68 0 x 185.43 Page 5 of 11 7. _____ 185 feet Practice Test Chapter 5 For Credit show all work 8. Find the width across the pool to the nearest foot. 200 feet 680 Adjacent = x 8. __________ 75 feet The angle of elevation is 680 The pool is the adjacent side = x The hypotenuse is 200 feet Cos = adjacent side / hypotenuse x cos 680 200 200 cos 680 x 74.9 9. If 00≥ x ≤ 3600, solve sin x 9. ___________ 3 2 Sin = y/r Y is negative in Quadrants III and IV 3 Sin 2400 and sin 3000 2 10. Given the triangle on the right, find B to the nearest B tenth of a degree if b = 12, and a c = 18. C Sin 2400 and 3000 10. ________ c b A For angle B b is the opposite side 41.80 For a right triangle c is the hypotenuse Sin-1(opposite/hypotenuse) = angle Sin-1(12/18)= 41.810 For exercise 11 and 12, round the numbers to the nearest tenth. 11. If ∆ABC, A= 470, B=580, b=23. Find a 11. ______________ Law of Sines: we have an acute triangle and a side angle pair. 19.8 feet a b c sin A sin B sin C a 23 sin 47 sin 58 sin 47 a 23 sin 47 sin 47 sin 58 23 sin 47 19.83 a sin 58 Page 6 of 11 Practice Test Chapter 5 For Credit show all work 12. If C=37.20, a =17.9 and b = 22.3, find the area of ∆ABC 12. _______________ 120.7 square units 1 K ab sin C 2 K .517.922.3sin 37.2 120.66 13. Determine the number of possible solutions if A=470, a = 2 13. ____________ and b=4. Law of Sines Ambiguous Case no solution 0 0 Case I because 47 < 90 a < b true 2 <4 a ? b sin a 2 ? 4 sin 470 2 < 2.9 using rules no solution For exercise 14 and 15, round the numbers to the nearest tenth. 14. In ∆ABC, A=670, b = 10 and c= 5, Find a 14. _________ Law of Cosines a= 9.3 2 2 2 a = b + c -2bc cosA a2 = 102 + 52 -2(10)(5)cos 670 a2 = 100 + 25 – 100 (cos 67) = 85.92 a = 85.92 = 9.26 15. In ∆ABC, a=8, b = 6 and c= 12, Find C 15. __________ Law of Cosines 117.30 c2 = a2+ b2 – 2ab cos C 122 = 82 + 62 -2(8)(6) cos C 144 = 64 + 36 -96 cos C 144- 64-36 = -96 cos C (144-64-36) / (-96) = cos C -0.45833 = cos C cos-1(-0.45833) = C = 117.27 16. If a=18, b = 22 and c =30, find the area of ∆ABC 16. _________ Using the law of cosines find an angle, 196.7 square units Then using the area formulas we can find the area. 1 a 2 b 2 c 2 2ab sin A K bc sin A 2 182 = 222 + 302—2(22)(30) cos A (182 - 222 - 302) = -1320 Cos A -1060 = -1320 Cos A Cos-1(-1060/-1320) = A = 36.57 36.60 Area = (.5)(22)(30)(sin 36.58) = 196.66 Page 7 of 11 Practice Test Chapter 5 For Credit show all work 17. Aviation A traffic helicopter is flying 1000 feet above the downtown area. To the right, the pilot sees the baseball stadium at an angle of depression of 630. To the left the pilot sees the football stadium at an angle of depression of 180. Find the distance between the two stadiums. Step 1. Draw the picture Step 2 Name Variables H = helicopter F = Football Stadium B = baseball stadium D = downtown Step 3 Write in constants Angle of depression is look down from the helicopter. 17. __________ H 180 630 1000 Feet F D B H 180 630 1000 Feet 180 F 630 D B Step 4 Write the question in terms of the drawing. Find the distance of FD plus DB There are two right triangle ∆FDH and ∆BDH FD is adjacent, DH is opposite B and DB is adjacent, DH is opposite B 1000 1000 tan 63 tan 18 DB FD 1000 1000 DB 509.52 FD 3077.68 tan 63 tan 18 3077.68 + 509.52 = 3587.2 Page 8 of 11 3587 feet Practice Test Chapter 5 For Credit show all work 18. Railways: The steepest railway in the world is the Katoomba Scenic Railway in Australia. The passenger car is pulled up the mountain by twin steel cables. It travels along the track 1020 feet to obtain a change of altitude of 647 feet. a) Find the angle of elevation b) How far does the train travel horizontally. Step 1. Draw the picture Step 2 Name Variables a= altitude b = horizontal Step 3 Write in constants 18 A) __________ 18 B) __________ B 647 feet C 1020 feet b A Step 4 Write the question in terms of the drawing Angle of elevation is looking up or A. 647 0 A sin 1 39.36 1020 2 2 647 b 1020 2 b 2 1020 2 647 2 621791 b 621791 788.53 Page 9 of 11 a) 39.40 b) 788 feet Practice Test Chapter 5 For Credit show all work 19. Engineering The escalator of St. Petersburg Metro in Russia has a vertical rise of 195.8 feet. If the angle of elevation of the escalator is 100 21’ 36”, find the length of the escalator? Step 1. Draw the picture Step 2 Name Variables c = length of the escalator Step 3 Write in constants 19. ________ B 195.8 feet x C Covert to decimal 10 21 / 60 36 / 3600 10.36 195.8 sin( 10.36) x 195.8 x 1088.79 sin 10.36 Page 10 of 11 A = 100 21’ 36” 1088.8 feet Practice Test Chapter 5 For Credit show all work 20. Architecture: A flagpole 40 feet high stands on top of the Wentworth Building. From the point in front of Bailey’s Drugstore, the angle of elevation for the top of the pole is 540 54’ and the angle to the bottom of the pole is 470 30’. How high is the building? Step 1. Draw the picture Step 2 Name Variables x = height of the building. d = distance to the building Step 3 Write in constants F 50 feet 20. _________ FAC= 540 54’ =54.9 BAC= 470 30’ =47.50 B x C Wentworth Building d A Bailey’s Drugstore 50 x 50 x 50 x d d tan FAC tan 54.9 x tan( BAC ) d x tan 47.5 d d tan 47.5 x tan( FAC) 50 x tan 47.5 x tan 54.9 50 tan 47.5 x tan 47.5 x tan 54.9 50 tan 47.5 x tan 54.9 x tan 47.5 50 tan 47.5 xtan 54.9 tan 47.5 50 tan 47.5 x 164.57 tan 54.9 tan 47.5 Page 11 of 11 165.6 feet