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Name: ________________________ Class: ___________________ Date: __________ ID: A Practice Trigonometry Test Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. If tan B = 1.0724 then the measure of ∠B, to the nearest degree, is a. 2° c. 43° b. 18° d. 47° ____ 2. In a. b. ____ 0.60 0.75 c. d. 1.20 1.33 3. Determine the measure of ∠C, to the nearest degree. a. b. ____ ABC, AB = 15 cm and AC = 9 cm. Determine the tangent ratio of ∠B, to the nearest hundredth. 19° 20° c. d. 21° 22° 4. A school soccer field measures 45 m by 65 m. To get home more quickly, you decide to walk along the diagonal of the field. What is the angle of your path, with respect to the 45-m side, to the nearest degree? a. 55° c. 2° b. 34° d. 1° 1 Name: ________________________ ID: A Use the diagram to answer the following question(s). ____ 5. Determine the length of y, to the nearest tenth of a metre. a. 5.7 m c. 7.0 m b. 6.3 m d. 8.2 m ____ 6. A wheelchair ramp is being built for the entrance to a school. If the ramp makes an angle of 4° with the ground and has a horizontal length of 6 m, determine the height of the ramp, to the nearest tenth of a metre. a. 1.1 m c. 0.4 m b. 0.8 m d. 0.3 m ____ 7. Evaluate sin 67°, to four decimal places. a. 2.3558 b. 0.9205 c. d. 0.8554 0.3907 ____ 8. Determine the value of sin 28°, to four decimal places. a. 0.4695 c. 0.8829 b. 0.5317 d. 0.9709 ____ 9. If cos A = 0.9903, then the measure of ∠A, to the nearest degree, is a. 45° c. 8° b. 82° d. 1° ____ 10. If cos B = 0.6561, then the measure of ∠B, to the nearest degree, is a. 49° c. 33° b. 41° d. 29° ____ 11. If sin A = 0.7986, determine the measure of ∠A, to the nearest degree. a. 1° c. 39° b. 7° d. 53° 2 Name: ________________________ ____ 12. In a. b. ID: A ABC, AC = 8 cm and BC = 11 cm. Determine the sine ratio of ∠B, rounded to the nearest thousandth. 0.588 0.728 c. d. 0.809 1.375 ____ 13. Write the cosine ratio of ∠A. a. b. AB AC BC AB ____ 14. Determine the value of cos 0°. a. –1 b. 0 ____ 15. In a. b. c. d. c. d. AC BC BC AC 1 undefined TUV, UV = 8 m, ∠U = 90°, and ∠T = 38°. Determine the measure of ∠V, to the nearest degree 42° c. 90° 52° d. 128° 3 Name: ________________________ ID: A Use the diagram to answer the following question(s). Kelly is flying a kite in a field. He lets out 40 m of his kite string, which makes an angle of 72° with the ground. ____ 16. Suppose the sun is shining directly above the kite. How far is the kite’s shadow from Kelly, to the nearest metre? a. 12 m c. 42 m b. 38 m d. 129 m 7 , what is the measure of ∠A, to the nearest degree? 8 62° c. 60° 61° d. 59° ____ 17. If sin A = a. b. ____ 18. Determine the length of x and the length of y, to the nearest tenth of a metre. a. b. x = 7.2 m and y = 9.7 m x = 9.6 m and y = 12.9 m c. d. 4 x = 9.6 m and y = 14.3 m x = 7.2 m and y = 10.8 m Name: ________________________ ID: A ____ 19. A window cleaner places a ladder that is 8 m long against a wall. The top of the ladder is 6 m above the ground. Determine the angle between the base of the ladder and the ground, to the nearest tenth of a degree. a. b. 36.9° 41.4° c. d. 48.5° 51.2° ____ 20. Which statement is incorrect? a. You can solve for the unknown side in any triangle, if you know the lengths of the other two sides, by using the Pythagorean theorem. b. The hypotenuse is the longest side in a right triangle. c. The hypotenuse is always opposite the 90° angle in a right triangle. d. The Pythagorean theorem applies to all right triangles. Completion Complete each statement. 1. The hypotenuse is the side _________________________ the right angle in a right triangle. 2. The tangent ratio of ∠A can be written as _________________________ . 3. The sine ratio of ∠A compares the length of the _________________________ side to the length of the hypotenuse. 4. The angle of elevation is the angle between the horizontal and the line of sight looking _________________________ to an object. 5. The angle between the horizontal and the line of sight looking up to an object is called the angle of _________________________ . 5 Name: ________________________ ID: A Matching Match the correct side or angle of PQR to each of the following definitions, descriptions, or expressions. A term may be used more than once or not at all. a. b. c. p p q p r f. q r g. –1 h. 0 d. Pythagorean theorem i. e. q j. ____ 1. the side opposite ∠Q ____ 2. r 2 = q 2 + p 2 1 pq 2 1 Match each term to the correct definition, description, or expression. A term may be used more than once or not at all. a. angle of elevation e. Pythagorean theorem b. angle of depression f. sine ratio c. cosine ratio g. tangent ratio d. hypotenuse h. trigonometry ____ 3. the angle between the horizontal and the line of sight looking down to an object ____ 4. in a right triangle with side lengths a, b, and c, where c is the hypotenuse, a 2 + b 2 = c 2 ____ 5. ____ 6. the side opposite the right angle in a right triangle ____ 7. length of side opposite an acute angle length of side adjacent to an acute angle length of side adjacent to an acute angle length of hypotenuse 6 Name: ________________________ ID: A Short Answer 1. A flight of stairs has a ratio of vertical distance to horizontal distance of 3 to 5. What angle does the flight of stairs make with the ground, to the nearest degree? 2. Denis is building a playhouse for his younger brother. The roof of the playhouse is shaped like an isosceles triangle. The diagonal trusses of the roof make an angle of 24° with the horizontal. The height of the roof is 3.5 m. How long are the diagonal trusses AB and AC, to the nearest tenth of a metre? 3. In XYZ, XY and XZ have equal lengths of 6 cm. YZ is 10 cm. Determine the measure of ∠X, to the nearest degree. 4. Triangle ABC is a right triangle. Side AB is 16 cm, and side AC is 18 cm. Using trigonometry, determine the length of BC, to the nearest tenth of a centimetre. 5. A telephone pole is secured with a guy-wire as shown in the diagram. The guy-wire makes an angle of 75° with the ground and is secured to the ground 6 m from the bottom of the pole. Determine the length of the wire, to the nearest tenth of a metre. 6. A ski jumper begins at the top of a ski jump hill that is 90 m high. The hill makes an angle of 30° with the horizontal. Determine the length of the hill, to the nearest metre. 7 Name: ________________________ ID: A 7. Khalil decides to go parasailing while on vacation. The flyer advertises that the maximum height reached during the trip will be 68 m. If the parasailing cable is 91 m long, what angle will the cable make with the horizontal when Khalil reaches the maximum height? Answer to the nearest degree. Problem 1. A submarine is travelling parallel to the surface of the ocean at a depth of 626 m. It begins a constant ascent in order to reach the surface after travelling a distance of 4420 m. The ascent takes 35 min. a) What angle of ascent, or angle of elevation, would the submarine need to make to reach the surface in 4420 m? Answer to the nearest degree. b) How far did the submarine travel horizontally during its ascent to the surface? Answer to the nearest metre. c) Determine the horizontal speed of the submarine during its ascent. Give your answer to the nearest metre per second. 2. The percent grade of a road is the ratio of the vertical rise of the road to the horizontal distance, expressed as a percent. A road in the mountains has a vertical rise of 40 m over 1 km of horizontal distance. a) Determine the percent grade of the road, to the nearest percent. b) What is the angle of elevation of the road, to the nearest degree? 3. A cell phone tower is supported by two guy wires, attached on opposite sides of the tower. One guy wire is attached to the top of the base of the tower at point A. The other is attached to the base at point E, at a height of 70 m above the ground. a) Demonstrate that ABC and EBD are similar triangles. b) Determine the height of the base of the tower, to the nearest metre. c) Determine the length of each guy-wire, to the nearest metre. 8 Name: ________________________ ID: A 4. José is sitting in a tree, so that his eyes are 3.2 m above the ground. When he looks down at an angle of depression of 43°, he can see his cat sitting in the yard. a) Draw a diagram of the situation. b) Determine the horizontal distance, to the nearest tenth of a metre, from the base of the tree to José’s cat. 5. Myriam is participating in a water-skiing competition. She goes over a water-ski ramp that is 4.0 m long. When she leaves the ramp, she is 1.7 m above the surface of the water. a) What is the horizontal length of the ramp along the surface of the water, to the nearest tenth of a metre? b) Determine the angle of elevation of the ramp, to the nearest degree. 6. A 100-m cable is attached to the top of a cell phone tower and is secured to the ground 45 m from the base of the tower. Determine the angle that the cable makes with the ground, to the nearest degree. 7. The string on Yuri’s kite is 45 m long and makes an angle of 55° with the ground. Yuri’s friend, Abdul, is standing directly below the kite. a) How far apart are Abdul and Yuri now, to the nearest tenth of a metre? b) Abdul runs away from Yuri, so that the angle of elevation between Abdul and the kite is 15°. How far apart are Abdul and Yuri, to the nearest tenth of a metre? 9 ID: A Practice Trigonometry Test Answer Section MULTIPLE CHOICE 1. ANS: NAT: KEY: 2. ANS: NAT: KEY: 3. ANS: NAT: KEY: 4. ANS: NAT: KEY: 5. ANS: NAT: KEY: 6. ANS: NAT: KEY: 7. ANS: NAT: 8. ANS: NAT: 9. ANS: NAT: KEY: 10. ANS: NAT: KEY: 11. ANS: NAT: KEY: 12. ANS: NAT: KEY: 13. ANS: NAT: KEY: 14. ANS: NAT: 15. ANS: NAT: KEY: D PTS: 1 DIF: B M4 TOP: The Tangent Ratio tangent ratio | determine an angle measure B PTS: 1 DIF: B M4 TOP: The Tangent Ratio tangent ratio | calculate a tangent ratio | right triangle C PTS: 1 DIF: B M4 TOP: The Tangent Ratio tangent ratio | determine an angle measure | right triangle A PTS: 1 DIF: C M4 TOP: Solving Right Triangles tangent ratio | determine an angle measure B PTS: 1 DIF: C M4 TOP: The Tangent Ratio tangent ratio | determine a distance using trigonometry C PTS: 1 DIF: B M4 TOP: Solving Right Triangles tangent ratio | determine a distance using trigonometry B PTS: 1 DIF: A M4 TOP: The Sine and Cosine Ratios A PTS: 1 DIF: A M4 TOP: The Sine and Cosine Ratios C PTS: 1 DIF: B M4 TOP: The Sine and Cosine Ratios cosine ratio | determine an angle measure A PTS: 1 DIF: B M4 TOP: The Sine and Cosine Ratios cosine ratio | determine an angle measure D PTS: 1 DIF: B M4 TOP: The Sine and Cosine Ratios sine ratio | determine an angle measure A PTS: 1 DIF: B M4 TOP: The Sine and Cosine Ratios sine ratio | calculate a sine ratio | right triangle A PTS: 1 DIF: A M4 TOP: The Sine and Cosine Ratios cosine ratio | write a cosine ratio | right triangle C PTS: 1 DIF: A M4 TOP: The Sine and Cosine Ratios B PTS: 1 DIF: A M4 TOP: The Sine and Cosine Ratios sine ratio | determine an angle measure 1 OBJ: Section 3.1 OBJ: Section 3.1 OBJ: Section 3.1 OBJ: Section 3.3 OBJ: Section 3.1 OBJ: Section 3.3 OBJ: KEY: OBJ: KEY: OBJ: Section 3.2 sine ratio | calculate a sine ratio Section 3.2 sine ratio | calculate a sine ratio Section 3.2 OBJ: Section 3.2 OBJ: Section 3.2 OBJ: Section 3.2 OBJ: Section 3.2 OBJ: Section 3.2 KEY: cosine ratio | calculate a cosine ratio OBJ: Section 3.2 ID: A 16. ANS: NAT: KEY: 17. ANS: NAT: KEY: 18. ANS: NAT: KEY: 19. ANS: NAT: KEY: 20. ANS: NAT: A PTS: 1 DIF: C OBJ: Section 3.3 M4 TOP: Solving Right Triangles cosine ratio | determine a distance using trigonometry | solve a right triangle B PTS: 1 DIF: A OBJ: Section 3.2 M4 TOP: The Sine and Cosine Ratios sine ratio | determine an angle measure D PTS: 1 DIF: C OBJ: Section 3.2 M4 TOP: The Sine and Cosine Ratios sine ratio | determine a distance using trigonometry | right triangle C PTS: 1 DIF: B OBJ: Section 3.3 M4 TOP: Solving Right Triangles sine ratio | determine a distance using trigonometry | solve a right triangle A PTS: 1 DIF: B OBJ: Section 3.3 M4 TOP: Solving Right Triangles KEY: Pythagorean theorem | hypotenuse COMPLETION 1. ANS: opposite PTS: 1 DIF: A TOP: The Tangent Ratio 2. ANS: Example: length of side opposite ∠A tanA = length of side adjacent to ∠A OBJ: Section 3.1 NAT: M4 KEY: right triangle | hypotenuse PTS: TOP: KEY: 3. ANS: 1 DIF: B OBJ: Section 3.1 NAT: M4 The Tangent Ratio tangent ratio | write a tangent ratio | primary trigonometric ratios opposite PTS: TOP: KEY: 4. ANS: 1 DIF: A OBJ: Section 3.2 NAT: M4 The Sine and Cosine Ratios sine ratio | define the sine ratio | opposite side | primary trigonometric ratios up PTS: 1 DIF: A TOP: Solving Right Triangles 5. ANS: elevation OBJ: Section 3.3 NAT: M4 KEY: angle of elevation PTS: 1 DIF: B TOP: Solving Right Triangles OBJ: Section 3.3 NAT: M4 KEY: angle of elevation MATCHING 1. ANS: E NAT: M4 PTS: 1 DIF: TOP: The Tangent Ratio 2 A OBJ: Section 3.1 KEY: label a right triangle | opposite side ID: A 2. ANS: D NAT: M4 3. ANS: NAT: 4. ANS: NAT: 5. ANS: NAT: KEY: 6. ANS: NAT: 7. ANS: NAT: KEY: PTS: 1 DIF: TOP: The Tangent Ratio A B PTS: 1 DIF: B M4 TOP: Solving Right Triangles E PTS: 1 DIF: A M4 TOP: Solving Right Triangles G PTS: 1 DIF: B M4 TOP: The Tangent Ratio tangent ratio | primary trigonometric ratios D PTS: 1 DIF: A M4 TOP: The Tangent Ratio C PTS: 1 DIF: B M4 TOP: The Sine and Cosine Ratios cosine ratio | primary trigonometric ratios OBJ: Section 3.1 KEY: Pythagorean theorem OBJ: KEY: OBJ: KEY: OBJ: Section 3.3 angle of depression Section 3.3 Pythagorean theorem Section 3.1 OBJ: Section 3.1 KEY: hypotenuse OBJ: Section 3.2 SHORT ANSWER 1. ANS: Let x be the angle the stairs make with the horizontal, in degrees. 3 tan x = 5 ÊÁ 3 ˆ˜ x = tan −1 ÁÁÁÁ ˜˜˜˜ Ë 5¯ x = 30.9638. . . The stairs make an angle of approximately 31° with the stairs. PTS: 1 DIF: TOP: The Tangent Ratio 2. ANS: AD sin B = AB sin 24° = AB = B OBJ: Section 3.1 NAT: M4 KEY: tangent ratio | determine an angle measure 3.5 AB 3.5 sin 24° AB = 8.6051. . . The trusses are each approximately 8.6 m long. PTS: 1 DIF: B OBJ: Section 3.2 NAT: M4 TOP: The Sine and Cosine Ratios KEY: sine ratio | determine a distance using trigonometry | isosceles triangle 3 ID: A 3. ANS: cos Z = side adjacent to ∠Z hypotenuse cos Z = ZW XZ cos Z = 5 6 cos Z = 0.8333 Z = cos −1(0.8333) Z = 33.5573. . . ∠Z measures approximately 34°. ∠Y is equal to ∠Z. 180 = ∠X + ∠Y + ∠Z 180 = ∠X + 34 + 34 180 = ∠X + 68 112 = ∠X ∠X measures approximately 112°. PTS: 1 DIF: C OBJ: Section 3.2 NAT: M4 TOP: The Sine and Cosine Ratios KEY: cosine ratio | determine an angle measure | isosceles triangle 4. ANS: side opposite ∠C sinC = hypotenuse sinC = AB AC sinC = 16 18 C = sin −1(0.8888) C = 62.7340. . . C = 63° BC cos 63° = 18 18(cos 63°) = BC 8.1718. . . = BC BC is approximately 8.2 cm. PTS: 1 DIF: C OBJ: Section 3.2 NAT: M4 TOP: The Sine and Cosine Ratios KEY: sine ratio | cosine ratio | determine a distance using trigonometry 4 ID: A 5. ANS: Let x represent the length of the wire, in metres. distance of wire from base cos 75° = length of wire cos 75° = x= 6 x 6 cos 75° x = 23.1822. . . The length of the wire is approximately 23.2 m. PTS: 1 DIF: B OBJ: Section 3.2 NAT: M4 TOP: The Sine and Cosine Ratios KEY: cosine ratio | determine a distance using trigonometry | right triangle 6. ANS: Let h represent the length of the hill, in metres. 90 sin30° = h h= 90 sin30° h = 180 The length of the ski jump hill is 180 m. PTS: 1 DIF: B OBJ: Section 3.2 NAT: M4 TOP: The Sine and Cosine Ratios KEY: sine ratio | determine a distance using trigonometry 7. ANS: Let x represent Khalil’s angle above the water, in degrees. height of Khalil above the water sinx = length of cable sinx = 68 91 sinx = 0.7473 x = sin −1 (0.7473) x = 48.3570. . . When Khalil is at the maximum height, the cable makes an angle of approximately 48°. PTS: 1 DIF: B TOP: The Sine and Cosine Ratios OBJ: Section 3.2 NAT: M4 KEY: sine ratio | determine an angle measure 5 ID: A PROBLEM 1. ANS: a) sin x = sin x = depth of submarine below surface distance travelled during ascent 626 4420 sin x = 0.1416 x = sin−1 (0.1416) x = 8.1421. . . The submarine’s angle of ascent is approximately 8°. b) Let d represent the distance travelled by the submarine, in metres.. d cos 8° = 4420 4420(cos 8°) = d 4376.9849. . . = d The submarine traveled approximately 4377 m horizontally. 1min = 60 s c) 35 min × 60 s = 2100 s 1min 4420 m = 2.1 2100 s During the ascent, the submarine travels at a horizontal speed of 2.1 m/s. PTS: 1 DIF: D OBJ: Section 3.3 NAT: M4 TOP: Solving Right Triangles KEY: sine ratio | cosine ratio | solve a right triangle | determine a distance using an angle of elevation | speed 6 ID: A 2. ANS: a) 1 km = 1000 m ÊÁ ˆ˜ vertical rise ˜˜ × 100 percent grade = ÁÁÁÁ ˜˜ horizontal distance Ë ¯ ÊÁ 40 ˆ˜ ˜˜ × 100 percent grade = ÁÁÁÁ ˜˜ 1000 Ë ¯ percent grade = 0.04 × 100 percent grade = 4 The percent grade of the road is 4%. b) Let θ represent the angle of elevation, in degrees. vertical rise tan θ = horizontal distance tan θ = 40 1000 θ = tan−1 (0.04) θ = 2.2906. . . The angle of elevation of the road is approximately 2°. PTS: 1 DIF: D TOP: Solving Right Triangles OBJ: Section 3.3 NAT: M4 KEY: tangent ratio | angle of elevation 7 ID: A 3. ANS: a) ∠C = ∠D = 70°; ∠ABC = ∠EBD = 90°; ∠A = ∠E = 20° AC corresponds to ED, BC corresponds to BD, and AB corresponds to EB, so the triangles are similar. height of tower tanC = b) distance of wire to base of tower tan 70° = AB 42 42(tan70°) = AB 115.3941... = AB The base of the tower is approximately 115 m tall. c) cos C = cos 70° = AC = distance from wire to base of tower length of wire 42 AC 42 cos 70° AC = 122.7998. . . The guy wire attached to the top of the base of the tower is approximately 123 m long. height of tower sinD = length of wire sin 70° = ED = 70 ED 70 sin 70° ED = 74.4924. . . The second guy wire is approximately 74 m long. PTS: 1 DIF: C OBJ: Section 3.3 NAT: M4 TOP: Solving Right Triangles KEY: tangent ratio | sine ratio | cosine ratio | solve a right triangle | similar triangles 8 ID: A 4. ANS: a) Example: b) Let d represent the distance from the cat to the base of the tree, in metres. distance from base of tree to cat tan43° = height of Jose above the ground tan43° = d 3.2 3.2(tan43°) = d 2.9840. . . = d The distance from the base of the tree to the cat is about 3.0 m. PTS: 1 DIF: A OBJ: Section 3.3 NAT: M4 TOP: Solving Right Triangles KEY: tangent ratio | determine a distance using an angle of depression 9 ID: A 5. ANS: a) Let x represent the length of the ramp along the surface of the water, in metres. Draw a diagram of the situation. x 2 + 1.7 2 = 4.0 2 x 2 = 16 − 2.89 x 2 = 13.11 x ≈ 3.62 The horizontal length of the ramp along the surface of the water is about 3.6 m. b) sinθ = 1.7 4.0 sinθ = 0.425 θ = 25.1507. . . θ ≈ 25 The angle of elevation of the ramp is about 25°. PTS: 1 DIF: B OBJ: Section 3.2 NAT: M4 TOP: The Sine and Cosine Ratios KEY: sine ratio | angle of elevation | Pythagorean theorem 6. ANS: Let x represent the angle, in degrees, that the cable makes with the ground. cos x = distance from cable to base of tower length of cable cos x = 45 100 x = cos −1 (0.45) x = 63.2563. . . The angle that the cable makes with the ground is approximately 63°. PTS: 1 DIF: A TOP: Solving Right Triangles OBJ: Section 3.3 NAT: M4 KEY: cosine ratio | determine an angle measure 10 ID: A 7. ANS: a) Let x represent the distance between Abdul and Yuri, in metres. distance between Abdul and Yuri cos 55° = length of kite string cos 55° = x 45 45(cos 55°) = x 25.8109. . . = x Abdul and Yuri are approximately 25.8 m apart. b) Let h represent the height of the kite above the ground, in metres. height of the kite above the ground sin55° = length of kite string sin55° = h 45 45(sin55°) = h 36.8618. . . = h Let x represent the horizontal distance between Abdul and the kite, in metres. height of the kite above the ground tan15° = horizontal distance between Abdul and the kite tan15° = x= 36.9 x 36.9 tan15° x = 137.713. . . Distance between Abdul and Yuri = 25.8 + 137.7 = 163.5 The distance between Abdul and Yuri is now approximately 163.5 m. PTS: 1 DIF: D OBJ: Section 3.2 NAT: M4 TOP: The Sine and Cosine Ratios KEY: cosine ratio | determine a distance using trigonometry | angle of elevation | solve a right triangle 11