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Answers to activity questions Check your skills 1. (a) Use Pythagoras’ Theorem to calculate x. h2 = a 2 + b2 x 2 = 302 + 102 30 cm x 2 = 900 + 100 x 2 = 1000 x x = 1000 x = 31.6cm 10cm (b) A rectangle has a length of 6.72m and width of 4.83m. Find the length of the diagonal (to 2 d.p.). Let the length of the diagonal be d. h2 = a 2 + b2 d 2 = 6.722 + 4.832 l na go D ia d 2 = 45.1584 + 23.3289 4.83m d 2 = 68.4873 d = 68.4873 d = 8.28cm 6.72m 2. (a) Calculate the value of x in this triangle. x opp adj x Tan 40° = 12.5 12.5 × Tan 40° = x Tanθ ° = 40° x = 10.49 (to 2 d.p.) 12.5m Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 1 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy (b) Calculate the value of x in this triangle. 7.3km 49° x (c) Calculate the value of θ in this triangle. 2m 1. θ θ = Sin −1 0.7083 θ = 45.10° or 45°5'48" 0.85m 3. opp hyp 7.3 Sin49° = x x × Sin49° = 7.3 7.3 x= Sin49° x = 9.67km opp Sinθ ° = hyp 0.85 Sinθ ° = 1.2 Sinθ ° = 0.7083 Sinθ ° = A kite on the end of a 30m string is flying at an angle of elevation of 72°. What is the height of the kite directly above the ground? opp hyp h Sin72° = 30 30 × Sin72° = h x = 28.53m 30 m Sinθ ° = h 72° Pythagoras’ Theorem 1. Use Pythagoras’ Theorem to find the missing length. (a) h2 = a 2 + b2 x 2 = 262 + 202 x 2 = 676 + 400 x 26 x 2 = 1076 x = 1076 x = 32.8 20 Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 2 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy (b) h2 = a2 + b2 x 182 = x 2 + 12.52 324 = x 2 + 156.25 324 − 156.25 = x 2 12.5 cm 18 cm x = 167.75 x = 12.95cm (c) Find the hypotenuse when a=6.1 and b=3.4 h2 = a2 + b2 x 2 = 6.12 + 3.42 x 2 = 37.21 + 11.56 x 2 = 48.77 (d) x = 48.77 x = 6.98 h2 = a2 + b2 Find the missing side given a=23.5 and h=40 402 = 23.52 + x 2 1600 = 552.25 + x 2 1600 − 552.25 = x 2 x = 1047.75 x = 32.37cm (e) Let the length of the linking side be l. 6m x 6m 8m (f) h2 = a 2 + b2 h2 = a2 + b2 l 2 = 62 + 82 x 2 = l 2 + 62 l 2 = 36 + 64 x 2 = 102 + 62 l 2 = 100 x 2 = 136 l = 100 l = 10m x = 136 x = 11.66m This is an equilateral triangle. Half the triangle to obtain a right angled triangle. h2 = a2 + b2 h 102 = h 2 + 52 10 mm Find the height of the triangle. Also calculate the area. Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning 100 = h 2 + 25 100 − 25 = x 2 x = 75 x = 8.66mm Page 3 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy (g) A orienteering participant runs 650m north an then turns and runs 1.4 km east. How far from the starting point is the runner? h 2 = a 2 + b2 650m = 0.65km x 2 = 2.3825 x 2 = 0.652 + 1.42 x 2 = 0.42252 + 1.962 x = 2.3825 x = 1.54km **A Real Challenge** A square has a diagonal of 20 cm, what is the side length? 20 cm (h) 202 = x 2 + x 2 400 = 2 x 2 400 x2 = 2 x = 200 x = 14.14cm x x 2. h2 = a 2 + b2 Do the following triangles contain a right angle? (a) 10, 24, 26 Remember the hypotenuse is the longest side. (b) n 7, 8, 10 2, 4.8, 5.2 a 2 + b 2 = 7 2 + 82 a 2 + b 2 = 22 + 4.82 a 2 + b 2 = 49 + 64 a 2 + b 2 = 4 + 23.04 a 2 + b 2 = 113 a 2 + b 2 = 27.04 a 2 + b 2 = 676 h 2 = 102 h 2 = 5.22 h 2 = 262 h 2 = 100 h 2 = 27.04 As a 2 + b 2 ≠ h 2 the triangle does not contain a right angle. As a 2 + b 2 = h 2 the triangle must contain a right angle. a 2 + b 2 = 102 + 242 a 2 + b 2 = 100 + 576 h 2 = 676 As a 2 + b 2 = h 2 the triangle must contain a right angle. (d) (c) 1.4, 4.8, 5 (e) 6, 6, 8 (f) 5, 6, 7 a 2 + b 2 = 1.42 + 4.82 a 2 + b2 = 62 + 62 a 2 + b 2 = 52 + 6 2 a 2 + b 2 = 1.96 + 23.04 a 2 + b 2 = 36 + 36 a 2 + b 2 = 25 + 36 a 2 + b 2 = 25 a 2 + b 2 = 72 a 2 + b 2 = 61 h 2 = 52 h 2 = 82 h2 = 72 h 2 = 25 h 2 = 64 h 2 = 49 Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 4 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy As a 2 + b 2 = h 2 the triangle does contain a right angle. As a 2 + b 2 ≠ h 2 the triangle does not contain a right angle. As a 2 + b 2 ≠ h 2 the triangle does not contain a right angle. The Trigonometric Ratios – Finding Sides 1. For the following triangles, identify the adjacent, opposite and hypotenuse for each triangle. (a) (b) (c) 5 θ° α° a 12 z y c 13 b 60° a – adjacent b – opposite c – hypotenuse x 5 – adjacent 12 – opposite 13 – hypotenuse x – adjacent y - opposite z - hypotenuse (e) (d) (f) 245 β° 4mm 45.3 γ° 25 h 450 34mm k 30mm δ° g 30mm – adjacent 34mm – opposite 45.34mm – hypotenuse g – adjacent h – opposite 25 – hypotenuse Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning 245 – adjacent 450 – opposite k – hypotenuse Page 5 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy 2. For the following triangles, find the value of the missing side(s). (a) (b) (c) 245 63° 22.5° 16m f 25 h b 58° opp hyp h Sin58° = 25 25 × Sin58° = h x = 21.2 Sinθ ° = f = 480.8 (e) (d) 14.7m (f) b e 24°45' m 33° adj hyp 14.7 Cos33° = m m × Cos33° = 14.7 14.7 m= Cos33° m = 17.5m Cosθ ° = opp adj b Tan 22.5° = 16 16 × Tan 22.5° = b b = 6.63m Tanθ ° = opp Tanθ ° = adj f Tan63° = 245 245 × Tan63° = f m 5c 72 adj hyp b Cos 24°45 ' = 725 725 × Cos 24°45 ' = b b = 658.4cm Cosθ ° = 68.52° 500m d opp hyp 500 Sin68.52° = e e × Sin68.52° = 500 500 e= Sin68.52° e = 537.3m Sinθ ° = Using Pythagoras e2 = 5002 + d 2 537.32 = 5002 + d 2 288691.3 − 250000 = d 2 d 2 = 38691.3 d = 38691.3 d = 196.7 Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 6 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy (g) (h) (i) 9° 36°36' 34 5m m h x h 57° 57° 62.5 km 24 cm Half the shape is a right angled triangle. adj Cosθ ° = hyp 345 Cos9° = x x × Cos9° = 345 345 x= Cos9° x = 349.3mm opp hyp 62.5 Sin36°36 ' = h h × Sin36°36 ' = 62.5 62.5 h= Sin36°36 ' h = 104.8km opp adj h Tan57° = 12 12 × Tan57° = h h = 18.48cm Sinθ ° = Tanθ ° = Area: 1 A = bh 2 A = 0.5 × 24 ×18.48 A = 221.76cm 2 3. 4. Express the following directions as a true bearing. (a) S15°E = 165°T (b) NE = 45°T (c) N45°W = 315°T (d) 10° W of N = 350°T (e) SSW = 202.5° (f) 215° = 215°T A person standing on top of a 25 m cliff sees a rower out to sea. The angle of depression of the boat is 15.3°. How far is the boat from the base of the cliff? 15.3° opp adj 25 Tan15.3° = d d × Tan15.3° = 25 Angle of Depression Tanθ ° = 25m by alternate angles = 15.3° d Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning 25 Tan15.3° d = 91.4m d= Page 7 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy 5. A student wishes to find the height of a building. From a distance of 50m on perfectly level ground, the angle of elevation to the top is 24.6°. Find the height of the building? opp adj h Tan 24.6° = 50 50 × Tan 24.6° = h h = 22.9m Tanθ ° = h 24.6° 50m 6. A plane is flying at altitude of 5000m. The pilot observes a boat at an angle of depression of 12°, calculate the horizontal distance which places the plane directly above the boat. opp adj 5000 Tan12° = d d × Tan12° = 5000 Tanθ ° = d 5000m 12° 5000 Tan12° d = 23523m or 23.5km d= 7. A walker decides to take a direct route to a landmark. They walk 1.7 km at a bearing of 78°T. How far did they walk in a northerly and easterly direction? N e n 78° adj hyp n Cos 78° = 1.7 1.7 × Cos 78° = n n = 0.353km or 353m opp Sinθ ° = hyp e Sin78° = 1.7 1.7 × Sin78° = e e = 1.663km Cosθ ° = km 7 . 1 Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 8 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy 8. Find the perimeter of this trapezium. opp adj h Tan51° = 16 16 × Tan51° = h h = 19.8cm Tanθ ° = 28 cm s h 51° adj hyp 16 Cos51° = s s × Cos51° = 16 16 s= Cos51° s = 25.4cm Cosθ ° = 44-28 = 16cm 44 cm Perimeter = 44 + h + 28 + s = 44 + 19.8 + 28 + 25.4 = 117.2cm or 1.172m 9. A kite is attached to a 45m line. On a windy day, the kite flies at an angle of elevation of 28°. Calculate the height of the kite above the ground. opp hyp h Sin 28° = 45 45 × Sin 28° = h h = 21.13m Sinθ ° = 45m h 28° 10. A plane flying at an altitude of 10 000m is flying away from a person. The angle of elevation of the plane is 76° when initially observed. After 1 minute 15 seconds, the plane is at an angle of elevation of 29°. Ignoring the height of the person, what is the speed of the plane in km/hr? Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 9 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy 10000m 76° dc di 29° df Initially, the plane is di away from the observer. After 1minute 15 seconds, the plane is d f away from the observer. The distance covered in 1 minute 15 seconds is d f − d i . opp adj 10000 Tan76° = di opp adj 10000 Tan 29° = df di × Tan76° = 10000 d f × Tan 29° = 10000 Tanθ ° = Tanθ ° = 10000 Tan76° di = 2493m d f − di = 18040 − 2493 = 15547 m or 15.547km 10000 Tan29° d f = 18040m di = df = 1 minute 15 seconds is 1.25 minutes. 1.25 minutes ÷ 60 =0.0208333 hours. distance covered (km) time taken(hr) 15.547 km = 0.0208333hr = 746km / hr (rounded) speed (km / hr ) = Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 10 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy The Trigonometric Ratios – Finding Angles 1. Find the size of the angle in the questions below to 2 d.p. (a) Sin θ = 0.2345 θ = 13.56° (b) Cos θ = 0.5736 θ = 55° (c) Tan θ = 2.4604 θ = 67.88° (d) Cos β = 0.63 β = 50.95° (e) Tan µ = 0.4998 µ = 26.56° (f) Sin σ = 0.9455 σ = 71° 2. Find the size of the angle in the questions below in degrees, minutes and seconds. (a) Sin θ = 0.2345 θ = 13°33’44” (b) Cos θ = 0.5736 θ = 54°59’54” (c) Tan θ = 2.4604 θ = 67°52’53” (d) Cos β = 0.63 β = 50°57’ (e) Tan μ = 0.4998 μ = 26°33’21” (f) Sin σ = 0.9455 σ = 70°59’48” 3. For the following triangles, find the size of the missing angle. (a) (b) (c) δ° 3.7m 25 cm 22 cm 4.45m 2800mm θ° α° opp Sinα ° = hyp 22 Sinα ° = 25 Sinα ° = 0.88 α ° = 61.64° or 61°38'33" (d) α° adj Cosθ ° = hyp 3.7 Cosθ ° = 4.45 Cosθ ° = 0.8315 θ = 33.75° or 33°45'3" (e) 1200mm opp adj 2800 Tanθ ° = 1200 Tanθ ° = 2.333 θ ° = 66.8° or 66°48'5" Tanθ ° = (f) 14.2 cm 2.4km 1800m β° 9.2 cm 3000mm θ° 2.4km = 2400 m θ° θ° 2900mm This is an isosceles triangle. Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 11 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy 4. Half the base is 1450mm. adj hyp 9.2 Cos β ° = 14.2 Cos β ° = 0.6479 β = 49.62° or 49°37'3" opp adj 1800 Tanθ ° = 2400 Tanθ ° = 0.75 θ ° = 36.9° or 36°52'12" Cos β ° = Tanθ ° = opp adj 3000 Tanθ ° = 1450 Tanθ ° = 2.0690 θ ° = 64.2° or 64°12'14" Tanθ ° = Complete the triangle below: Calculate x using Pythagoras. 2 α° β° cm cm 3 5 2 h = a +b x 2 x 2 = 52 + 32 x 2 = 25 + 9 x 2 = 34 x = 34 x = 5.83 Calculating α opp adj 3 Tanα ° = 5 Tanα ° = 0.6 α ° = 30.96° or 30°57'50" Tanα ° = Calculating β β = 90 − α β = 90 − 30.96° β = 59.04° or β = 90 − 30°57 '50" β = 59°2 '10" 5. A ship sails 400 km north and then 800 km west. How far is it away from the port it started from and at what true bearing? Find d using Pythagoras. N h 2 = a 2 + b2 400 km 800 km d θ° d 2 = 4002 + 8002 d 2 = 160000 + 640000 d 2 = 800000 d = 800000 d = 894.4km Finding the true bearing opp adj 800 Tanθ ° = 400 Tanθ ° = 2 θ ° = 63.4° or 63°26 '6" Tanθ ° = The ship is located 894km away at a true bearing of 63.4°. Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 12 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy 6. It is recommended that ramps for wheelchairs have a rise to run ratio of 1:15. What angle would the ramp surface make to the horizontal if a ramp follows this specification? Units are not required to answer this question. 1 θ° 15 The surface of the ramp should make an angle of about 4° to the horizontal to be wheelchair friendly. 7. opp adj 1 Tanθ ° = 15 Tanθ ° = 0.0666 θ ° = 3.8° or 3°48'51" Tanθ ° = Two flag poles are in a park on a level surface 30m apart. One pole is 10m tall and the other is 19m tall. What is the angle of elevation of the top of the tall pole from the top of the smaller pole? opp adj 9 Tanθ ° = 30 Tanθ ° = 0.3 θ ° = 16.7° or 16°41'57" Tanθ ° = 19-10=9m θ° 30m 19m 10m 8. A 10m antenna is supported by guide wires on four sides. Each wire is attached to the ground 6 metres from the base of the antenna. Find the length of each wire h2 = a 2 + b2 2 2 d = 6 + 10 l 2 d 2 = 36 + 100 10m d 2 = 136 θ° d = 136 d = 11.66m What angle does the wire make with the ground? opp adj 10 Tanθ ° = 6 Tanθ ° = 1.666 θ ° = 59.04° or 59°2 '10" Tanθ ° = 6m Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 13 [last edited on 22 July 2015] CRICOS Provider: 01241G Centre for Teaching and Learning Numeracy 9. Calculate the pitch of this roof. (Assuming symmetry) This is an isosceles triangle. Half the large triangle is a right angled triangle. opp adj 2.5 Tanθ ° = 4.1 Tanθ ° = 0.6098 θ ° = 31.4° or 31°22 ' 23" 2.5m Tanθ ° = θ° 8.2m 10. In ∆ABC : ∠A = 90°, a = 16.9, b = 6.5, calculate∠B . The first step is to draw the diagram to match the information given. B θ° 16.9 A 6.5 opp hyp 6.5 Sinθ ° = 16.9 Sinθ ° = 0.3846 ∠B = θ ° = 22.62° or 22°37'12" Sinθ ° = C Centre for Teaching and Learning | Academic Practice | Academic Skills T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning Page 14 [last edited on 22 July 2015] CRICOS Provider: 01241G