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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 11 · Surveying & Navigation

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.5.74
2.22.6∘22.6^\circ
3.13
4.32\dfrac{\sqrt{3}}{2}
5.24 cm2^2
6.60∘60^\circ
7.52.2
8.65∘65^\circ
9.asin⁡A=bsin⁡B\dfrac{a}{\sin A} = \dfrac{b}{\sin B}
10.a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A
11.π3\dfrac{\pi}{3}
12.150∘150^\circ
13.12\dfrac{1}{2}
14.12\dfrac{1}{2}
15.1
16.12 cm
17.36 cm2^2
18.12.6 cm
19.25.1 cm2^2
20.Both equal 22\dfrac{\sqrt{2}}{2}.
21.3
22.π\pi
23.Max 8, Min 2
24.5
25.y=4sin⁡x+1y = 4\sin x + 1
26.12
27.π4\dfrac{\pi}{4} units right
28.Sinusoidal (sine or cosine).
29.Amplitude 5; mean 9
30.Amplitude 3; period π\pi; no vertical shift.
Silver
31.11.7
32.48.3∘48.3^\circ
33.7.00
34.44.4∘44.4^\circ
35.(a) 52.2 m2^2; (b) BC≈13.0BC \approx 13.0 m
36.31.231.2 m
37.6.43 cm
38.North 25.0 km; East 43.3 km
39.636\sqrt{3}
40.≈299\approx 299 m
41.−32-\dfrac{\sqrt{3}}{2}
42.−12-\dfrac{1}{2}
43.x=2π3x = \dfrac{2\pi}{3} or 4π3\dfrac{4\pi}{3}
44.45\dfrac{4}{5}
45.34\dfrac{3}{4}
46.30.6 cm
47.32\dfrac{\sqrt{3}}{2}
48.−sin⁡θ-\sin\theta and cos⁡θ\cos\theta.
49.x=2π3x = \dfrac{2\pi}{3} or 5π3\dfrac{5\pi}{3}
50.(a) 4π4\pi cm; (b) 18π18\pi cm2^2
51.(a) 3; (b) 12; (c) 7
52.t≈1.39t \approx 1.39 hours
53.d(0)=5d(0) = 5 m; max 8 m
54.t=3t = 3 hours
55.h(t)=15−12cos⁡ ⁣(πt2)h(t) = 15 - 12\cos\!\left(\dfrac{\pi t}{2}\right)
56.18°C
57.Period 4 s; amplitude 6
58.A=5A = 5, C=7C = 7
59.y=3cos⁡ ⁣(πx4)+1y = 3\cos\!\left(\dfrac{\pi x}{4}\right) + 1
60.18 = height of centre above ground; 15 = radius of the wheel.
Gold
61.42.0∘42.0^\circ
62.∠PAB=90∘\angle PAB = 90^\circ; PB≈94.3PB \approx 94.3 km
63.≈87.3∘\approx 87.3^\circ
64.≈11.4\approx 11.4
65.≈45.6∘\approx 45.6^\circ
66.50=52\sqrt{50} = 5\sqrt{2} cm
67.≈25.1∘\approx 25.1^\circ
68.230∘230^\circ
69.B≈52.1∘B \approx 52.1^\circ or 127.9∘127.9^\circ
70.≈9.32\approx 9.32 km
71.5π4\dfrac{5\pi}{4}
72.−1+32-\dfrac{1 + \sqrt{3}}{2}
73.x=π4,3π4,5π4,7π4x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}
74.≈61.4\approx 61.4 cm2^2
75.LHS =sin⁡2θsin⁡θ=sin⁡θ= \dfrac{\sin^2\theta}{\sin\theta} = \sin\theta.
76.cos⁡θ=−35\cos\theta = -\dfrac{3}{5}, tan⁡θ=43\tan\theta = \dfrac{4}{3}
77.x=π12,5π12,13π12,17π12x = \dfrac{\pi}{12}, \dfrac{5\pi}{12}, \dfrac{13\pi}{12}, \dfrac{17\pi}{12}
78.6−24\dfrac{\sqrt{6} - \sqrt{2}}{4}
79.105∘105^\circ
80.x=π6,5π6,3π2x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}
81.t≈1.31t \approx 1.31 min
82.t≈1.39t \approx 1.39 h or t≈4.61t \approx 4.61 h
83.L(t)=−3cos⁡ ⁣(πt6)+12L(t) = -3\cos\!\left(\dfrac{\pi t}{6}\right) + 12
84.(a) t=15t = 15 h (3pm); (b) 28°C
85.≈5.22\approx 5.22 hours
86.A=4A = 4, C=2C = 2
87.y=−4cos⁡ ⁣(πt4)+6y = -4\cos\!\left(\dfrac{\pi t}{4}\right) + 6
88.y=cos⁡(x−π/2)y = \cos(x - \pi/2)
89.8
90.x(t)=15cos⁡(πt)x(t) = 15 \cos(\pi t)
Platinum
91.(a) 85∘85^\circ; (b) CL≈8.24CL \approx 8.24 km
92.≈32.0\approx 32.0 m
93.≈10.5\approx 10.5 km
94.Case 1: B≈52.1∘B \approx 52.1^\circ, C≈92.9∘C \approx 92.9^\circ, c≈13.9c \approx 13.9. Case 2: B≈127.9∘B \approx 127.9^\circ, C≈17.1∘C \approx 17.1^\circ, c≈4.10c \approx 4.10.
95.(a) 272\sqrt{7}; (b) 636\sqrt{3}
96.≈46.7∘\approx 46.7^\circ
97.≈49.3\approx 49.3 cm2^2
98.≈49.4\approx 49.4 km
99.≈232\approx 232 m
100.≈911\approx 911 km
101.r=12r = 12; segment ≈13.7\approx 13.7 cm2^2
102.θ=32\theta = \dfrac{3}{2} rad ≈85.9∘\approx 85.9^\circ
103.See working.
104.x=π6,5π6,3π2x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}
105.r=5r = 5 cm; max area 25 cm2^2
106.6+24\dfrac{\sqrt{6} + \sqrt{2}}{4}
107.θ=π3\theta = \dfrac{\pi}{3}; segment =6π−93≈3.26= 6\pi - 9\sqrt{3} \approx 3.26 cm2^2
108.x=0,π2,2πx = 0, \dfrac{\pi}{2}, 2\pi
109.11π2\dfrac{11\pi}{2} cm2^2
110.See working.
111.≈1.38\approx 1.38 min
112.t≈0.671t \approx 0.671 h
113.(a) 8; (b) π\pi; (c) x(t)=8cos⁡(πt)x(t) = 8\cos(\pi t)
114.L(t)=−3.5cos⁡(πt/6)+12L(t) = -3.5\cos(\pi t/6) + 12; L(3)=12L(3) = 12 h
115.(a) 60 Hz; (b) t=1240t = \dfrac{1}{240} s ≈4.17\approx 4.17 ms
116.y(t)=4sin⁡ ⁣(πt4)y(t) = 4\sin\!\left(\dfrac{\pi t}{4}\right)
117.Periodic with period 2π2\pi (least common period of components).
118.A=4A = 4, B=π6B = \dfrac{\pi}{6}, C=2C = 2
119.13\dfrac{1}{3} (8 hours per day)
120.10 W

Pack B — Answers

Bronze
1.7.71
2.16.3∘16.3^\circ
3.17
4.12\dfrac{1}{2}
5.30 cm2^2
6.120∘120^\circ
7.53.6
8.70∘70^\circ
9.bsin⁡B=csin⁡C\dfrac{b}{\sin B} = \dfrac{c}{\sin C}
10.b2=a2+c2−2accos⁡Bb^2 = a^2 + c^2 - 2ac \cos B
11.3π4\dfrac{3\pi}{4}
12.120∘120^\circ
13.22\dfrac{\sqrt{2}}{2}
14.32\dfrac{\sqrt{3}}{2}
15.3\sqrt{3}
16.12 cm
17.30 cm2^2
18.6.28 cm
19.52.4 cm2^2
20.sin⁡60∘=32\sin 60^\circ = \dfrac{\sqrt{3}}{2}; cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2}.
21.7
22.6
23.Max 9, Min 5
24.7
25.y=6sin⁡(2x)+2y = 6\sin(2x) + 2
26.8
27.π3\dfrac{\pi}{3} units left
28.Sinusoidal.
29.Amplitude 7; mean 15
30.Amplitude 4; period 6; shifted down 1.
Silver
31.10.3
32.62.1∘62.1^\circ
33.10.2
34.49.5∘49.5^\circ
35.(a) 74.8; (b) BC≈14.4BC \approx 14.4
36.37.337.3 m
37.15.3 cm
38.North −51.4-51.4 km (i.e. 51.4 km south); East 61.3 km
39.1534\dfrac{15\sqrt{3}}{4}
40.≈275\approx 275 m
41.−12-\dfrac{1}{2}
42.−22-\dfrac{\sqrt{2}}{2}
43.x=π4x = \dfrac{\pi}{4} or 3π4\dfrac{3\pi}{4}
44.1213\dfrac{12}{13}
45.512\dfrac{5}{12}
46.21.8 cm
47.32\dfrac{\sqrt{3}}{2}
48.−tan⁡θ-\tan\theta; since tan⁡(−θ)=sin⁡(−θ)cos⁡(−θ)=−sin⁡θcos⁡θ\tan(-\theta) = \frac{\sin(-\theta)}{\cos(-\theta)} = \frac{-\sin\theta}{\cos\theta}.
49.x=π4x = \dfrac{\pi}{4} or 5π4\dfrac{5\pi}{4}
50.(a) 10π10\pi cm; (b) 60π60\pi cm2^2
51.(a) 5; (b) 8; (c) 12
52.t≈1.33t \approx 1.33 hours
53.d(0)=6d(0) = 6 m; max 8 m
54.t=9t = 9 hours
55.h(t)=12−10cos⁡ ⁣(πt3)h(t) = 12 - 10\cos\!\left(\dfrac{\pi t}{3}\right)
56.23°C
57.Period 4 s; amplitude 4
58.A=5A = 5, C=2C = 2
59.y=−4cos⁡ ⁣(πx3)+2y = -4\cos\!\left(\dfrac{\pi x}{3}\right) + 2
60.22 = mean (average) temperature; 6 = amplitude of temperature variation.
Gold
61.75.4∘75.4^\circ
62.∠PAB=85∘\angle PAB = 85^\circ; PB≈77.5PB \approx 77.5 km
63.≈97.9∘\approx 97.9^\circ
64.≈12.0\approx 12.0
65.≈47.7∘\approx 47.7^\circ
66.77\sqrt{77} cm
67.≈27.1∘\approx 27.1^\circ
68.300∘300^\circ
69.B≈45.6∘B \approx 45.6^\circ or 134.4∘134.4^\circ
70.≈11.5\approx 11.5 km
71.5π3\dfrac{5\pi}{3}
72.−1+32-\dfrac{1 + \sqrt{3}}{2}
73.x=π3,2π3,4π3,5π3x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}
74.≈52.8\approx 52.8 cm2^2
75.LHS =(1−cos⁡2θ)−cos⁡2θ=1−2cos⁡2θ= (1 - \cos^2\theta) - \cos^2\theta = 1 - 2\cos^2\theta = RHS.
76.sin⁡θ=1213\sin\theta = \dfrac{12}{13}, tan⁡θ=−125\tan\theta = -\dfrac{12}{5}
77.x=π8,7π8,9π8,15π8x = \dfrac{\pi}{8}, \dfrac{7\pi}{8}, \dfrac{9\pi}{8}, \dfrac{15\pi}{8}
78.6+24\dfrac{\sqrt{6} + \sqrt{2}}{4}
79.110∘110^\circ
80.x=0,2π3,4π3x = 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}
81.t≈1.59t \approx 1.59 min
82.t≈1.33t \approx 1.33 h or t≈6.67t \approx 6.67 h
83.L(t)=−4cos⁡ ⁣(πt6)+12L(t) = -4\cos\!\left(\dfrac{\pi t}{6}\right) + 12
84.(a) t=14t = 14 h (2pm); (b) 24°C
85.≈8\approx 8 hours
86.A=5A = 5, C=7C = 7
87.y=−5cos⁡ ⁣(πt3)+9y = -5\cos\!\left(\dfrac{\pi t}{3}\right) + 9
88.y=sin⁡(x+π/2)y = \sin(x + \pi/2)
89.12
90.x(t)=20cos⁡ ⁣(2πt3)x(t) = 20 \cos\!\left(\dfrac{2\pi t}{3}\right)
Platinum
91.≈288∘\approx 288^\circ
92.≈30.1\approx 30.1 m
93.≈7.02\approx 7.02 km
94.Case 1: B≈45.6∘B \approx 45.6^\circ, C≈104.4∘C \approx 104.4^\circ, c≈13.6c \approx 13.6. Case 2: B≈134.4∘B \approx 134.4^\circ, C≈15.6∘C \approx 15.6^\circ, c≈3.77c \approx 3.77.
95.(a) 77; (b) 10310\sqrt{3}
96.≈48.5∘\approx 48.5^\circ
97.≈35.3\approx 35.3 cm2^2
98.≈49.0\approx 49.0 km
99.≈290\approx 290 m
100.≈645\approx 645 km
101.r≈13.07r \approx 13.07; segment ≈4.85\approx 4.85 cm2^2
102.θ=2\theta = 2 rad ≈114.6∘\approx 114.6^\circ
103.See working.
104.x=π6,5π6,π2,3π2x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{\pi}{2}, \dfrac{3\pi}{2}
105.r=7.5r = 7.5 cm; max area 56.25 cm2^2
106.2−64\dfrac{\sqrt{2} - \sqrt{6}}{4}
107.θ=π3\theta = \dfrac{\pi}{3}; segment =32π3−163≈5.79= \dfrac{32\pi}{3} - 16\sqrt{3} \approx 5.79 cm2^2
108.x=π4,5π4x = \dfrac{\pi}{4}, \dfrac{5\pi}{4}
109.2π2\pi cm2^2
110.See working: sin⁡θcos⁡θ=0\sin\theta \cos\theta = 0.
111.≈0.823\approx 0.823 min
112.t≈1.39t \approx 1.39 h
113.(a) 12; (b) π/2\pi/2; (c) x(t)=12cos⁡(πt/2)x(t) = 12\cos(\pi t/2)
114.L(t)=−3.5cos⁡(πt/6)+12.5L(t) = -3.5\cos(\pi t/6) + 12.5; L(3)=12.5L(3) = 12.5 h
115.(a) 50 Hz; (b) t=1200t = \dfrac{1}{200} s =5= 5 ms
116.y(t)=3sin⁡ ⁣(πt3)y(t) = 3\sin\!\left(\dfrac{\pi t}{3}\right)
117.Periodic with period 2π2\pi.
118.A=6A = 6, B=π4B = \dfrac{\pi}{4}, C=5C = 5
119.≈1/6\approx 1/6 (about 4 h)
120.18 W

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 52.2 m2^2. (b) 13.0 m. (c) 42.0∘42.0^\circ.
Full working
(a) Area =12(12)(9)sin⁡75∘≈52.2= \frac{1}{2}(12)(9)\sin 75^\circ \approx 52.2. (b) BC2=144+81−2(12)(9)cos⁡75∘≈169.1BC^2 = 144 + 81 - 2(12)(9)\cos 75^\circ \approx 169.1, BC≈13.0BC \approx 13.0. (c) Sine rule: sin⁡B=9sin⁡75∘13.0≈0.669\sin B = \frac{9 \sin 75^\circ}{13.0} \approx 0.669, B≈42.0∘B \approx 42.0^\circ.
2Problem 2
Answer
(a) BC=27BC = 2\sqrt{7}. (b) Area =63= 6\sqrt{3}.
Full working
(a) BC2=16+36−24=28⇒BC=27BC^2 = 16 + 36 - 24 = 28 \Rightarrow BC = 2\sqrt{7}. (b) 12(4)(6)⋅32=63\frac{1}{2}(4)(6) \cdot \frac{\sqrt{3}}{2} = 6\sqrt{3}.
3Problem 3
Answer
(a) At S1S_1: 45∘45^\circ; at S2S_2: 65∘65^\circ; at FF: 70∘70^\circ. (b) S2F≈60.2S_2 F \approx 60.2 m. (c) Height ≈32.0\approx 32.0 m.
Full working
(a) Bearings: at S1S_1, FF is 45∘45^\circ east of north → ∠FS1S2=45∘\angle FS_1S_2 = 45^\circ. At S2S_2, S1S_1 is south (bearing 180180) and FF is 115115; difference =65∘= 65^\circ. (b) ∠F=70∘\angle F = 70^\circ. Sine rule: S2F=80sin⁡45∘sin⁡70∘≈60.2S_2F = \frac{80 \sin 45^\circ}{\sin 70^\circ} \approx 60.2 m. (c) Height =60.2tan⁡28∘≈32.0= 60.2 \tan 28^\circ \approx 32.0.
4Problem 4
Answer
(a) Interior angle at VV is 85∘85^\circ. (b) CL≈8.24CL \approx 8.24 km. (c) ≈288∘\approx 288^\circ.
Full working
(a) Turn =145−50=95∘= 145 - 50 = 95^\circ, interior =85∘= 85^\circ. (b) CL2=25+49−70cos⁡85∘≈67.9CL^2 = 25 + 49 - 70\cos 85^\circ \approx 67.9, CL≈8.24CL \approx 8.24. (c) Sine rule for ∠VCL\angle VCL: sin⁡∠VCL7=sin⁡85∘8.24\frac{\sin\angle VCL}{7} = \frac{\sin 85^\circ}{8.24}, ∠VCL≈57.9∘\angle VCL \approx 57.9^\circ. Bearing LL from CC: 107.9∘107.9^\circ. Return: +180∘=287.9∘+180^\circ = 287.9^\circ.
5Problem 5
Answer
(a) Leg 1: E 5.91, N 1.04; Leg 2: E 4.50, N −7.79-7.79; Leg 3: E −3.76-3.76, N −1.37-1.37. (b) Total E ≈6.65\approx 6.65; total N ≈−8.12\approx -8.12. (c) ≈10.5\approx 10.5 km.
Full working
(a) East =dsin⁡θ= d\sin\theta; North =dcos⁡θ= d\cos\theta. (b) Add component-wise. (c) Distance =E2+N2= \sqrt{E^2 + N^2}.
6Problem 6
Answer
(a) B≈52.1∘B \approx 52.1^\circ or 127.9∘127.9^\circ. (b) Case 1: C≈92.9∘C \approx 92.9^\circ, c≈13.9c \approx 13.9. Case 2: C≈17.1∘C \approx 17.1^\circ, c≈4.10c \approx 4.10.
Full working
(a) sin⁡B=11sin⁡35∘8≈0.789\sin B = \frac{11 \sin 35^\circ}{8} \approx 0.789. Two valid angles. (b) Use the sine rule again for cc.
7Problem 7
Answer
(a) 525\sqrt{2} cm. (b) ≈25.1∘\approx 25.1^\circ. (c) ≈45.0∘\approx 45.0^\circ.
Full working
(a) 25+16+9=50=52\sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}. (b) Base diagonal 41\sqrt{41}; tan⁡θ=341\tan\theta = \frac{3}{\sqrt{41}}, θ≈25.1∘\theta \approx 25.1^\circ. (c) cos⁡θ=552=12\cos\theta = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}, θ=45∘\theta = 45^\circ.
8Problem 8
Answer
(a) tan⁡35∘=h/d\tan 35^\circ = h/d and tan⁡25∘=h/(d+200)\tan 25^\circ = h/(d + 200). (b) d≈399d \approx 399 m; h≈280h \approx 280 m.
Full working
(a) Two right triangles with same height. (b) h=dtan⁡35∘=(d+200)tan⁡25∘h = d \tan 35^\circ = (d + 200)\tan 25^\circ; subtract to get d(tan⁡35∘−tan⁡25∘)=200tan⁡25∘d(\tan 35^\circ - \tan 25^\circ) = 200 \tan 25^\circ. d≈399.4d \approx 399.4; h≈279.6h \approx 279.6 m. To 3 s.f.: d≈399d \approx 399, h≈280h \approx 280.
9Problem 9
Answer
(a) A:800A: 800 km; B:700B: 700 km. (b) 70∘70^\circ. (c) ≈864\approx 864 km.
Full working
(c) d2=8002+7002−2(800)(700)cos⁡70∘d^2 = 800^2 + 700^2 - 2(800)(700)\cos 70^\circ. With cos⁡70∘≈0.342\cos 70^\circ \approx 0.342: d2≈1 130 000−383 040≈747 000d^2 \approx 1\,130\,000 - 383\,040 \approx 747\,000, d≈864d \approx 864 km.
10Problem 10
Answer
(a) Largest angle is opposite side 11; ≈85.9∘\approx 85.9^\circ. (b) ≈30.6\approx 30.6 unit2^2. (c) Acute (all angles <90∘< 90^\circ).
Full working
(a) cos⁡C=49+81−121126=9126≈0.0714\cos C = \frac{49 + 81 - 121}{126} = \frac{9}{126} \approx 0.0714, C≈85.9∘C \approx 85.9^\circ. (b) Use 12(7)(9)sin⁡85.9∘≈31.4\frac{1}{2}(7)(9)\sin 85.9^\circ \approx 31.4 or Heron: s=13.5s = 13.5; A=13.5⋅6.5⋅4.5⋅2.5=986.7≈31.4A = \sqrt{13.5 \cdot 6.5 \cdot 4.5 \cdot 2.5} = \sqrt{986.7} \approx 31.4 unit2^2. (c) Largest angle <90∘< 90^\circ, so triangle is acute.
11Problem 11
Answer
(a) ≈47.7∘\approx 47.7^\circ. (b) ≈13.4\approx 13.4 m.
Full working
(a) 12(12)(18)sin⁡θ=80⇒sin⁡θ=160216≈0.7407\frac{1}{2}(12)(18)\sin\theta = 80 \Rightarrow \sin\theta = \frac{160}{216} \approx 0.7407, θ≈47.8∘\theta \approx 47.8^\circ. (b) Cosine rule: c2=144+324−432cos⁡47.8∘≈468−290.5≈177.5c^2 = 144 + 324 - 432\cos 47.8^\circ \approx 468 - 290.5 \approx 177.5, c≈13.3c \approx 13.3 m.
12Problem 12
Answer
(a) 45∘45^\circ. (b) AT≈184AT \approx 184 m; BT≈205BT \approx 205 m. (c) ≈178\approx 178 m.
Full working
(a) ∠ATB=180−75−60=45∘\angle ATB = 180 - 75 - 60 = 45^\circ. (b) AT=150sin⁡60∘sin⁡45∘≈183.7AT = \frac{150 \sin 60^\circ}{\sin 45^\circ} \approx 183.7; BT=150sin⁡75∘sin⁡45∘≈204.9BT = \frac{150 \sin 75^\circ}{\sin 45^\circ} \approx 204.9. (c) Drop a perpendicular from TT to ABAB: height =ATsin⁡(∠TAB)=183.7sin⁡75∘≈177.4= AT \sin(\angle TAB) = 183.7 \sin 75^\circ \approx 177.4 m.
13Problem 13
Answer
(a) 32\frac{\sqrt{3}}{2}. (b) 32\frac{\sqrt{3}}{2}. (c) 1. (d) 12\frac{1}{2}.
Full working
From the 30-60-90 and 45-45-90 special triangles, with π6=30∘\frac{\pi}{6} = 30^\circ.
14Problem 14
Answer
−32-\dfrac{\sqrt{3}}{2}.
Full working
5π6\frac{5\pi}{6} is in QII (between π2\frac{\pi}{2} and π\pi). Reference angle: π−5π6=π6\pi - \frac{5\pi}{6} = \frac{\pi}{6}. Cosine is negative in QII: cos⁡5π6=−cos⁡π6=−32\cos\frac{5\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}.
15Problem 15
Answer
(a) 4π≈12.64\pi \approx 12.6 cm. (b) 18π≈56.518\pi \approx 56.5 cm2^2. (c) 18+4π≈30.618 + 4\pi \approx 30.6 cm.
Full working
(a) s=rθ=4πs = r\theta = 4\pi. (b) A=12r2θ=18πA = \frac{1}{2}r^2\theta = 18\pi. (c) Perimeter =2r+s=18+4π= 2r + s = 18 + 4\pi.
16Problem 16
Answer
(a) x=2π3x = \frac{2\pi}{3} or 4π3\frac{4\pi}{3}. (b) x=45∘x = 45^\circ or 135∘135^\circ. (c) x=2π3x = \frac{2\pi}{3} or 5π3\frac{5\pi}{3}.
Full working
(a) cos⁡x=−12\cos x = -\frac{1}{2}, ref π3\frac{\pi}{3}, cos negative in QII, QIII. (b) Sin positive in QI, QII. (c) Tan negative in QII, QIV; ref π3\frac{\pi}{3}.
17Problem 17
Answer
(a) −45-\frac{4}{5}. (b) −34-\frac{3}{4}. (c) −1225-\frac{12}{25}.
Full working
(a) cos⁡2θ=1−925=1625\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}. QII → cosine negative: −45-\frac{4}{5}. (b) tan⁡=sin⁡/cos⁡=−34\tan = \sin/\cos = -\frac{3}{4}. (c) Multiply.
18Problem 18
Answer
(a) Major sector 200π3≈209.4\frac{200\pi}{3} \approx 209.4 cm2^2. (b) 253≈43.325\sqrt{3} \approx 43.3 cm2^2. (c) Minor segment ≈61.4\approx 61.4 cm2^2.
Full working
(a) Minor sector =100π3= \frac{100\pi}{3}; major sector =πr2−minor=200π3= \pi r^2 - \text{minor} = \frac{200\pi}{3}. (b) Triangle =12r2sin⁡θ=253= \frac{1}{2}r^2\sin\theta = 25\sqrt{3}. (c) Minor segment =100π3−253≈61.4= \frac{100\pi}{3} - 25\sqrt{3} \approx 61.4.
19Problem 19
Answer
See working.
Full working
Cross-multiplying: (1−cos⁡θ)(1+cos⁡θ)=sin⁡2θ(1 - \cos\theta)(1 + \cos\theta) = \sin^2\theta. LHS =1−cos⁡2θ=sin⁡2θ= 1 - \cos^2\theta = \sin^2\theta by Pythagorean identity. Hence the identity holds, provided sin⁡θ≠0\sin\theta \neq 0 (LHS denominator) and 1+cos⁡θ≠01 + \cos\theta \neq 0 (RHS denominator), i.e. θ≠nπ\theta \neq n\pi.
20Problem 20
Answer
x=π6,5π6,3π2x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}.
Full working
Let u=sin⁡xu = \sin x: 2u2+u−1=(2u−1)(u+1)=02u^2 + u - 1 = (2u - 1)(u + 1) = 0. u=12u = \frac{1}{2} → x=π6,5π6x = \frac{\pi}{6}, \frac{5\pi}{6}. u=−1u = -1 → x=3π2x = \frac{3\pi}{2}.
21Problem 21
Answer
(a) 6+24\frac{\sqrt{6} + \sqrt{2}}{4}. (b) 2−64\frac{\sqrt{2} - \sqrt{6}}{4}.
Full working
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B with A=π4A = \frac{\pi}{4}, B=π3B = \frac{\pi}{3}. cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B.
22Problem 22
Answer
(a) θ=24−2rr\theta = \dfrac{24 - 2r}{r}. (b) A=r(12−r)=12r−r2A = r(12 - r) = 12r - r^2. (c) r=6r = 6, max area 36 cm2^2.
Full working
(a) Perimeter =2r+rθ= 2r + r\theta, so θ=24−2rr\theta = \frac{24 - 2r}{r}. (b) A=12r2θ=12r2⋅24−2rr=r(12−r)A = \frac{1}{2}r^2\theta = \frac{1}{2}r^2 \cdot \frac{24 - 2r}{r} = r(12 - r). (c) Quadratic A=−r2+12rA = -r^2 + 12r, max at r=6r = 6, A=36A = 36.
23Problem 23
Answer
(a) 33π≈10433\pi \approx 104 cm2^2. (b) 11π2≈17.3\frac{11\pi}{2} \approx 17.3 cm2^2. (c) 11π/3+6≈17.511\pi/3 + 6 \approx 17.5 cm. Wait — recompute: outer arc =7⋅π3=7π3= 7 \cdot \frac{\pi}{3} = \frac{7\pi}{3}; inner arc =4⋅π3=4π3= 4 \cdot \frac{\pi}{3} = \frac{4\pi}{3}; two radial edges total 2⋅3=62 \cdot 3 = 6. Total perimeter =7π+4π3+6=11π3+6≈17.5= \frac{7\pi + 4\pi}{3} + 6 = \frac{11\pi}{3} + 6 \approx 17.5 cm.
Full working
(a) Outer area =49π= 49\pi; inner area =16π= 16\pi; difference =33π= 33\pi. (b) Annular sector area =12(49−16)π3=33π6=11π2= \frac{1}{2}(49 - 16)\frac{\pi}{3} = \frac{33\pi}{6} = \frac{11\pi}{2}.
24Problem 24
Answer
(a) Both sides give 0. (b) Two more solutions visible, one positive, one negative. (c) x≈1.90x \approx 1.90. (d) The equation mixes a transcendental (sin⁡\sin) and a polynomial — there is no algebraic closed-form.
Full working
(a) sin⁡0=0\sin 0 = 0 and 0/2=00/2 = 0. (b) The line y=x/2y = x/2 has gradient 0.50.5; intersects sin⁡\sin once positive, once negative (by symmetry). (c) Use Newton's method or the GDC: x≈1.8955x \approx 1.8955. (d) Equations involving combinations of polynomial and transcendental functions are generally not solvable in closed form — only special cases (e.g. sin⁡x=0\sin x = 0) have explicit solutions.
25Problem 25
Answer
(a) Amplitude 3 m; period 12 h; mean 5 m. (b) Max 8 m; min 2 m. (c) t≈1.39t \approx 1.39 h (about 01:24).
Full working
(a) Read off A,B,CA, B, C. (b) Max =5+3= 5 + 3; min =5−3= 5 - 3. (c) sin⁡(πt/6)=2/3⇒t=6arcsin⁡(2/3)/π≈1.39\sin(\pi t/6) = 2/3 \Rightarrow t = 6 \arcsin(2/3) / \pi \approx 1.39.
26Problem 26
Answer
(a) See working. (b) Max 33 m, min 3 m. (c) t≈1.31t \approx 1.31 min. (d) ≈ 1.38 min.
Full working
(a) Mean 18; amplitude 15; period 4 ⇒ B=π/2B = \pi/2. Use −cos⁡-\cos so h(0)=18−15=3h(0) = 18 - 15 = 3 (min). (b) Max 33, min 3. (c) Solve 18−15cos⁡(πt/2)=25⇒cos⁡(πt/2)=−7/15⇒t≈1.3118 - 15\cos(\pi t/2) = 25 \Rightarrow \cos(\pi t/2) = -7/15 \Rightarrow t \approx 1.31. (d) By symmetry, above 25 m between t≈1.31t \approx 1.31 and t≈2.69t \approx 2.69, total ≈1.38\approx 1.38 min.
27Problem 27
Answer
(a) 3. (b) 12. (c) Max 7; min 1. (d) Starts at (0,4)(0, 4) (mean), rises to max 7 at x=3x = 3.
Full working
(a)–(c) Read off. (d) First quarter-period =3= 3; first max at x=3,y=7x = 3, y = 7. Mean =4= 4 at x=0,6,12x = 0, 6, 12.
28Problem 28
Answer
(a) Amplitude 3; mean 12. (b) L(t)=−3cos⁡(πt/6)+12L(t) = -3\cos(\pi t / 6) + 12. (c) 12 h. (d) t≈4.09t \approx 4.09 months.
Full working
(a) Read from max/min. (b) Min at t=0t = 0 → use −cos⁡-\cos. (c) cos⁡(π/2)=0\cos(\pi/2) = 0, L(3)=12L(3) = 12. (d) cos⁡(πt/6)=−1/3⇒πt/6=arccos⁡(−1/3)≈1.911\cos(\pi t/6) = -1/3 \Rightarrow \pi t/6 = \arccos(-1/3) \approx 1.911, t≈3.65t \approx 3.65 months. (Recompute: arccos⁡(−1/3)≈1.9106\arccos(-1/3) \approx 1.9106; t≈6⋅1.9106/π≈3.65t \approx 6 \cdot 1.9106 / \pi \approx 3.65.)
29Problem 29
Answer
(a) Amplitude 10; period 2. (b) 10, 0, −10-10. (c) t=0.5t = 0.5 or 1.51.5. (d) Max speed 10π≈31.410\pi \approx 31.4 cm/s.
Full working
(a) Read off. (b) Direct evaluation. (c) cos⁡(πt)=0⇒πt=π/2,3π/2\cos(\pi t) = 0 \Rightarrow \pi t = \pi/2, 3\pi/2. (d) v(t)=−10πsin⁡(πt)v(t) = -10\pi \sin(\pi t); max magnitude 10π10\pi.
30Problem 30
Answer
(a) y=cos⁡(x−π/2)y = \cos(x - \pi/2). (b) y=sin⁡(x+π/2)y = \sin(x + \pi/2). (c) y=cos⁡(x−π)y = \cos(x - \pi).
Full working
(a) Sine lags cosine by π/2\pi/2. (b) Cosine leads sine by π/2\pi/2. (c) −cos⁡x=cos⁡(x−π)-\cos x = \cos(x - \pi).
31Problem 31
Answer
(a) 4 s. (b) Amplitude 4; mean 5. (c) y(t)=4sin⁡(πt/2)+5y(t) = 4\sin(\pi t / 2) + 5.
Full working
(a) Pattern repeats every 4 s. (b) Max 9, min 1, mean 5. (c) Period 4 → B=π/2B = \pi/2. Sin choice because mean-and-rising at t=0t = 0.
32Problem 32
Answer
(a) ≈3.83\approx 3.83 months. (b) From t≈4.09t \approx 4.09 to t≈7.91t \approx 7.91 months. (c) Sketch.
Full working
L>13⇔cos⁡(πt/6)<−1/3⇔πt/6∈(arccos⁡(−1/3),2π−arccos⁡(−1/3))L > 13 \Leftrightarrow \cos(\pi t/6) < -1/3 \Leftrightarrow \pi t/6 \in (\arccos(-1/3), 2\pi - \arccos(-1/3)). With arccos⁡(−1/3)≈1.911\arccos(-1/3) \approx 1.911, πt/6∈(1.911,4.372)\pi t/6 \in (1.911, 4.372), so t∈(3.65,8.35)t \in (3.65, 8.35) months. Length ≈4.70\approx 4.70 months.
33Problem 33
Answer
(a) Periodic, period 2π2\pi. (b) Not periodic — the xx term grows. (c) Periodic, period 2π2\pi. (d) Periodic, period π\pi (since sin⁡xcos⁡x=12sin⁡2x\sin x \cos x = \frac{1}{2}\sin 2x).
Full working
(a) Both terms have period 2π2\pi. (b) Adding a non-periodic term breaks periodicity. (c) Periods π\pi and 2π/32\pi/3; LCM = 2π2\pi. (d) Identity sin⁡xcos⁡x=12sin⁡2x\sin x \cos x = \frac{1}{2}\sin 2x has period π\pi.
34Problem 34
Answer
(a) Amplitude 0.5 Pa; frequency 440 Hz. (b) Period ≈2.27\approx 2.27 ms. (c) P(0)=0P(0) = 0; P(1/1760)=0.5P(1/1760) = 0.5. (d) The pitch A4 (concert A).
Full working
(a) Read off. (b) T=1/f=1/440T = 1/f = 1/440 s ≈2.27\approx 2.27 ms. (c) At t=0t = 0: sin⁡0=0\sin 0 = 0. At t=1/1760t = 1/1760: 2π⋅440⋅1/1760=π/2⇒sin⁡=12\pi \cdot 440 \cdot 1/1760 = \pi/2 \Rightarrow \sin = 1, P=0.5P = 0.5. (d) 440 Hz is the international tuning standard for concert A.
35Problem 35
Answer
(a) Amplitude 4, period 12, mean 7. (b) d(0)=7d(0) = 7. (c) t=6t = 6 h. (d) ≈4.40\approx 4.40 h.
Full working
(a) Read off. (b) sin⁡0=0\sin 0 = 0. (c) Next zero of sin⁡(πt/6)\sin(\pi t/6) after t=0t = 0 is πt/6=π⇒t=6\pi t/6 = \pi \Rightarrow t = 6. (d) sin⁡(πt/6)<−1/2\sin(\pi t/6) < -1/2 in uu-space: (7π/6,11π/6)(7\pi/6, 11\pi/6). Convert: t∈(7,11)t \in (7, 11). Length 4 h. (Adjust: 4 h exactly.)
36Problem 36
Answer
(a) Data oscillates: high–low–high–even higher. The change is broadly periodic. (b) Amplitude ≈170\approx 170; mean ≈410\approx 410; period ≈12\approx 12. (c) Approximately P(t)=170sin⁡ ⁣(π(t−7)6)+410P(t) = 170\sin\!\left(\dfrac{\pi(t - 7)}{6}\right) + 410. (d) P(12)≈410+170sin⁡(5π/6)=410+85=495P(12) \approx 410 + 170\sin(5\pi/6) = 410 + 85 = 495.
Full working
(a) Periodicity is plausible for ecological populations with seasonal effects. (b) From extremes: max 580, min 240; amplitude =(580−240)/2=170= (580-240)/2 = 170, mean (580+240)/2=410(580 + 240)/2 = 410. Period ≈ 12 (high at t=10t = 10, next high would be at t=22t = 22, but only one full period in data; estimate). (c) Choose phase so model fits the observed minimum at t=4t = 4. (d) Substitute t=12t = 12.