Special Angles 1 – Worksheet MCR3U Jensen 1) a) Draw a right triangle that has one angle measuring 30°. Label the sides using lengths 3, 2, and 1. b) Identify the adjacent and opposite sides relative to the 30° angle. c) Redraw the triangle and identify the adjacent and opposite sides relative to the 60° angle. 2) a) Draw a right triangle that has one angle measuring 45°. Label the sides using the lengths 1, 1, and 2. b) Identify the adjacent and opposite sides relative to one of the 45° angles. 3) State the exact values a) π ππ60° b) πππ 30° c) π‘ππ45° d) πππ 45° 4) Determine the exact value of each trig expression. a) π ππ30° × π‘ππ60° − πππ 30° !"#!"° b) 1 − !"#!"° 5) Using exact values, show that π ππ! π + πππ ! π = 1 for each angle a) π = 30° b) π = 45° c) π = 60° 6) Using the appropriate special triangle, determine π if 0° ≤ π ≤ 90° a) π πππ = ! ! b) 3 π‘πππ = 1 c) 2 2 πππ π = 2 d) 2 πππ π = 3 7) A 5 meter stepladder propped against a classroom wall forms an angle of 30° with the wall. Exactly how far is the top of the ladder from the floor? Express your answer in radical form. 8) Simplify the following expressions by rationalizing the denominator a) b) c) ! ! ! ! ! ! !!! ! !! ! Answers 1) a) 2) a) ! b) ! b) ! 3) a) ! b) ! c) 1 d) ! 4) a) 0 d) 0 ! ! ! ! ! ! 5) a) ! + ! = 1 b) ! + 6) a) 60° b) 30° c) 45° d) 30° 7) ! ! ! 8) a) c) ! ! ! = 1 c) m ! !" ! b) 2 15 c) −4 + 4 2 + 5 3 − 5 6 ! ! ! + ! ! ! = 1 Special Angles 2 – Worksheet MCR3U Jensen 1) Sketch each angle in standard position and determine the related acute angle. a) 1350 e) -2250 i) 1050 b) 2100 f) -3300 j) -1630 c) 3150 g) 1500 k) -1410 d) -300 h) -1200 l) -2800 2) State the value of each ratio exactly. a) sin 225° b) cos 240° c) sin 270° d) tan 300° e) cos 180° 3) Point P (-­β9, 4) is on the terminal arm of an angle in standard position. a) Sketch the principal angle, θ. b) What is the measure of the related acute angle to the nearest degree? c) What is the measure of θ to the nearest degree? 4) Point P (7, -­β24) is on the terminal arm of an angle in standard position. a) Sketch the principal angle, θ. b) What is the measure of the related acute angle to the nearest degree? c) What is the measure of θ to the nearest degree? Answers 1) a) 450 b) 300 c) 450 d) 300 e) 450 f) 300 g) 300 h) 600 i) 750 j) 170 k) 390 l) 800 2) a) − 1 2 b) − 1 c) –1 2 3) b) 240 c) 1560 4) b) 740 c) 2860 d) − 3 e) –1 4.2 Related and Co-­βTerminal Angles – Worksheet MCR3U Jensen 1. For each point given on the terminal arm of the angle, determine the exact primary values for the trigonometric ratios of the angle. a) b) 2. One of the primary trigonometric ratios for an angle is given, as well as the quadrant in which each angle is located. Find the other two trigonometric ratios of the angle. 3 a) sin A = , first quadrant 4 2 b) cos B = − , second quadrant 3 9 c) tan C = , third quadrant 11 ! d) tan D =− , fourth quadrant !" 3. Determine two other angles that have the same trigonometric ratios as each given angle. a) sin 60° b) cos 230° c) tan 200° d) sin 150° 4. Determine any three positive angles that are co-­βterminal with 150°. c) d) Answers 12 . 37 a) Which quadrants can ∠F be in? b) Find the coordinates of a point on the terminal arm of the angle in each quadrant. c) If you are also told that the sine of the angle is negative, in which quadrant is ∠F? d) Write the other primary trigonometric ratios for ∠F in the quadrant identified in part c. 5. Consider ∠F such that cos F = 3 1. a) sin θ = 1 , cos θ = 10 b) sin θ = 4 5 , cos θ = − 5 c) sin θ = − 61 5 d) sin θ = − , tan θ = 3 10 3 5 , tan θ = − 6 , cos θ = − 34 3 , tan θ = 61 3 , cos θ = 4 34 , tan θ = − 5 6 5 3 7 2. a) cos A = b) sin B = 4 5 , tan B = − 3 d) sin D = − 7 5 c) sin C = − 3 , tan A = 11 2 , cos C = -­β 202 5 13 9 202 !" , cos D = !" 3. Answers may vary. Sample answers: a) sin 120°, cos 30° b) cos 130°, cos 590° c) tan 20°, tan 560° d) sin 30°, cos 60° 4. Answers may vary. Sample answer: 510°, 870° 5. a) the first and fourth quadrants b) first quadrant: (12, 35); fourth quadrant: (12, −35) c) fourth quadrant d) sin F = − 35 37 , tan F = − !" !" 4.3 Reciprocal Trig Ratios – Worksheet MCR3U Jensen 1) Determine the measure of each angle, to the nearest degree, if the angles are in the first quadrant. a) cot A = 7 7 b) sec B = 3 11 c) csc C = 8 2) Determine the exact expressions for the six trigonometric ratios for 315°. Hint: Draw a diagram of the angle in standard position. Then use special triangles to determine the exact values. 3) Find the measure, to the nearest degree, of an angle in the first quadrant that satisfies each ratio. If there is no such angle, explain why. ! a) π ππ π΄ = ! ! b) πππ π΅ = c) π‘ππ πΆ = ! !" ! ! d) ππ π π· = ! ! e) π ππ πΈ = ! ! f) πππ‘ πΉ = ! ! g) ππ π πΊ = − ! ! h) π ππ π» = ! 4) Determine two angles between 0° and 360° that have a secant of − 2. 5) Determine two angles between 0° and 360° that have a cotangent of −1. 6) Determine two angles between 0° and 360° that have a cosecant of 3. 7) Each point lies on the terminal arm of an angle in standard position. Determine exact expressions for the six trigonometric ratios for the angle. a) P(-­β5, 12) b) T(9, 40) c) V(5, -­β3) Answers 1) a) A = 8° b) B = 65° 2) sin 315° = − 1 2 c) C = 47° 1 2 sec315° = 2 cos315° = tan 315° = −1 cot 315° = −1 csc315° = − 2 3) a) 42° b) 53° c) 67° d) 63° e) 41° f) 53° 4) 135° and 225° 5) 135° and 315° 6) 19° and 161° 12 7) a) sin θ = 13 cscθ = 13 12 −5 13 5 =− 13 13 secθ = −5 13 =− 5 12 −5 12 =− 5 −5 cot θ = 12 5 =− 12 9 41 41 sec θ = 9 40 9 9 cot θ = 40 cosθ = tan θ = b) sin θ = cscθ = 40 41 41 40 cos θ = −3 34 c) 3 =− 34 34 cscθ = −3 sin θ = 34 =− 3 cosθ = tan θ = 5 34 5 −3 5 =− 3 cot θ = 34 secθ = 5 4.4 Problems in 2-­βDimensions Worksheet MCR3U Jensen 1) For each of the following, select the most appropriate trigonometric tool among primary trig ratios, sine law, and cosine law. Then use that tool to find the indicated unknown quantity. Round answer to 1 decimal place. a) In βπ΄π΅πΆ, ∠π΄ = 90°, ∠π΅ = 39°, and π = 10 cm. Determine π. b) In βπππ , ∠π = 35°, ∠π = 65°, and π = 3 m. Determine π. c) In βπ·πΈπΉ, ∠π· = 60°, ∠πΉ = 65°, and π = 12 cm. Determine π. d) In βπππ, ∠π = 42°, π¦ = 25 km and, and π§ = 20 km. Determine π₯ . 2) The shadow of a tree that is 12 meters tall measures 9 meters in length. Determine the angle of elevation of the sun. 3) There is a water hazard between a golfer’s ball and the green. The golfer has two choices. He can hit the ball alongside the water hazard to a point left of the green and play the next shot from there. Or, he can hit directly over the water hazard to the green. The golfer can usually hit an approach shot at least 60 meters. Should he attempt the direct shot, or go around the hazard? 4) Yolanda flies her ultra-­βlight airplane due east for 100 km. She turns right through an angle of 130°, and flies a second leg. Then, she turns right 110° and returns to her starting point. a) Represent the flight path using an appropriate diagram, labeling all information. b) Determine the total length of the flight, to the nearest km. 5) Find an exact expression for the distance between Bill and Nadia’s houses. 16 14 12 10 8 6 4 2 6) A radio antenna is stabilized by two guy wires. One guy wire is 100 m in length and is attached to the top of the antenna. The wire makes an angle of 60° with the ground. One end of the second guy wire is attached to the ground at the same point as the first guy wire. The other end is attached to the antenna such that the wire makes an angle of 45° with the ground. Determine an exact expression for the distance between the points where the two guy wires are attached to the antenna. 7) A Ferris wheel has a radius of 20 m, with 10 cars spaced around the circumference at equal distances. If the cars are numbered in order, how far is it directly from the first car to the fifth car? Answers 1) a) 6.3 cm b) 5.2 m c) 12.6 cm d) 16.8 km 2) 53° 3) 67 m. He should go for the direct shot. 4) a) A B b) 274 km 100 km Start 60° 50° 130° 70° ° 110 C 5) 20 181 + 90 2 m. 6) 50 3 − 1 m. 7) approximately 38 m. 4.5 Problems in 3-­βDimensions Worksheet MCR3U Jensen 1) The bases on a baseball diamond are 27.4 m apart. The pitcher pitches, and the batter hits a fly ball straight up 15 m. What is the maximum angle of elevation of the ball, to the nearest degree, as seen by the pitcher if he is standing at the center of the diamond? 2) A square-­βbased tent has the cross-­βsectional shape shown. The side wall goes up at an angle of elevation of 60° for 2 m, then continues at an angle of elevation of 30° for another 2 m to the peak. Determine an exact value for the height of the tent. 3) The Great Pyramid of Cheops at Giza in Egypt has a square base of side length 230 m. The angle of elevation of one triangular face is 52°. Determine the measure of the angle π between the height and one of the edges where two triangular faces meet. 4) A box in the shape of a square-­βbased prism has a base of side length 10 cm and a height of 20 cm. A rod dropped into the box lies exactly from one of the bottom corners to the opposite top corner. Determine the angle between the rod and the vertical edge of the box, rounded to one decimal place. 5) Jodi and Leanna are on top of a cliff at point D, 100 m above the base. They decide to race to a picnic table at A, where lunch is waiting. Jodi runs, at a constant speed of 5 m/s, down the hill from D to C and then directly to A. Leanna climbs down the cliff to B at a constant rate of 1 m/s, and then runs as fast as Jodi to A. Who reaches lunch first? 6) Ranjeet praks his car in a lot on the corner of Park Lane and Main Street. He walks 80 m east to First Aveunue, turns 30° to the left, and follows First Avenue for 100 m to the Metro Building, where he takes the elevator to his office on the 15th floor. Each floor in the building is 4 m in height. From his offie window, Ranjeet can see his car in the lot. How far is Ranjeet from his car, in a direct line? Answers 1) 38° 2) 3 + 1 m 3) approximately 48° 4) approximately 35° 5) Jodi takes 190 seconds, Joanna takes 213 seconds. Jodi reaches lunch first. 6) approximately 184 m The Ambiguous Case of Sine Worksheet MCR3U Jensen 1) In βπ΄π΅πΆ, π = 13 cm, π = 21 cm, and ∠π΄ = 29°. Draw possible diagrams that match the given measurements. Then calculate the length of side π . 2) In βπ΄π΅πΆ, π = 5.9 m, π = 7.8 m, and ∠π΄ = 36°. Draw possible diagrams that match the given measurements. Then calculate the length of side π . 3) In βπ΄π΅πΆ, π = 2.4 cm, π = 3.2 cm, and ∠π΄ = 28°. Determine two possible measures for ∠πΆ and for the length of side π. 4) In βπ·πΈπΉ, π = 3 cm, π = 5 cm, and ∠π· = 30°. Determine two possible measures for ∠πΈ and for the length of side π. 5) Two ships, S and T, are 120 km apart when they pick up a distress call from a yacht. Ship T estimates that the yacht is 70 km away and that the angle between the line from T to S and the line from S to the yacht is 28°. What are two possible distances, to the nearest tenth of a km, from ship S to the yacht? Answers 1) 2) 3) 4) 5) 26.5 cm or 10.3 cm 10 m or 2.6 m ∠πΆ = 39° and π = 4.7 cm; ∠πΆ = 141° and π = 1.0 cm ∠πΈ = 56° and π = 6 cm; ∠πΈ = 124° and π = 2.7 cm 147.5 km or 64.4 km 4.6 Trig Identities Worksheet #1 MCR3U Jensen 1) Prove each identity a) π πππ = πππ π π‘πππ b) ππ ππ = π πππ πππ‘π c) πππ π = π πππ πππ‘π d) π πππ = ππ ππ π‘πππ 2) Prove each identity a) 1 + ππ ππ = ππ ππ(1 + π πππ) b) πππ‘π π πππ π πππ = 1 c) πππ π π πππ − 1 = 1 − πππ π d) 1 + π πππ = π πππ(1 + ππ ππ) 3) Prove that 1 − π ππ! π = π πππ πππ π πππ‘π 4) Prove that ππ π ! π = πππ‘ ! π + 1 !"#$ !!!"#$ 5) Prove that !!!"#$ = !"#$ !"#$ !"!" ! 6) Prove that !!!"#$ + !!!"#$ = !"#$ 7) Prove that ππ π ! π πππ ! π = ππ π ! π − 1 8) Prove that π‘ππ! π − π ππ! π = π ππ! π π‘ππ! π 2.6 Trig Identities Worksheet #2 MCR3U Jensen Prove each of the following identities 1) πππ π × π‘πππ = π πππ 2) !!!"#! ! !"#$ = !!!"#! ! !"#$ !"# ! ! 3) πππ‘ ! π = !!!"#! ! 4) !"!# !"#$ = πππ‘π 5) π πππ + πππ π ! = 1 + 2 π πππ πππ π 6) 2 π ππ! π − 1 = π ππ! π − πππ ! π 7) ! !"#! ! ! ! + !"#! ! = !"#! ! !"#! ! 8) πππ ! π = π ππ! π + 2 πππ ! π − 1 9) π‘πππ = π‘ππ! π × πππ‘π 10) π ππ ! π + ππ π ! π = π ππ ! π × ππ π ! π 11) !!!"#$ + !!!"#$ = 2 π ππ ! π 12) π‘ππ! π − π ππ! π = π ππ! π × π‘ππ! π 13) 14) !"#$!! + !"#$!! = 0 15) !"!#!! + !"!#!! = 2 π ππ ! π ! ! !!!!"#$ !"#$ !"#$!!"#$ = π πππ + πππ π !"#$!! !"#$!! !"!# !"!#