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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 9 · Architecture

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.5 cm
2.525\sqrt{2}
3.252\sqrt{5}
4.Yes — right-angled
5.6
6.Rational: 5, 0.3‾0.\overline{3}, 4=2\sqrt{4} = 2. Irrational: 2,π\sqrt{2}, \pi.
7.≈ 7.1
8.15 cm
9.828\sqrt{2}
10.8 cm
11.160 cm³
12.96 cm²
13.90π90\pi cm³
14.24 cm²
15.35 cm²
16.108 cm³
17.144π144\pi cm³
18.36π36\pi cm³
19.108 cm²
20.112π112\pi cm²
21.sin⁡θ=3/5\sin \theta = 3/5
22.Opp = 10sin⁡30°=510 \sin 30° = 5; adj = 10cos⁡30°=5310 \cos 30° = 5\sqrt{3}; hyp = 10
23.3/2\sqrt{3}/2
24.tan⁡θ=5/12\tan \theta = 5/12
25.6
26.4 cm
27.1
28.5
29.1/31/\sqrt{3} or 3/3\sqrt{3}/3
30.Third side = 5; sin⁡θ=5/13\sin \theta = 5/13
Silver
31.21\sqrt{21}
32.5
33.323\sqrt{2}
34.2132\sqrt{13}
35.5
36.32=42\sqrt{32} = 4\sqrt{2}
37.13
38.24 cm
39.7+437 + 4\sqrt{3}
40.232\sqrt{3}
41.600 cm³
42.224 cm²
43.144π144\pi cm²
44.h=V/(πr2)h = V/(\pi r^2)
45.9 cm
46.96 cm³
47.20=25\sqrt{20} = 2\sqrt{5} cm
48.(a) (250/3)π(250/3)\pi (b) 50π50\pi
49.0.4 m
50.(a) 0.25 m³ (b) 250 000 cm³
51.θ≈24.6°\theta \approx 24.6°
52.5.74 cm
53.4.62 cm
54.(a) 10 (b) ≈ 53.1°
55.≈ 32°
56.8.66
57.5sin⁡60°≈4.335 \sin 60° \approx 4.33 m
58.(a) 28.1° (b) 17
59.≈ 68.7 m
60.sin⁡−1(5/13)≈22.6°\sin^{-1}(5/13) \approx 22.6°
Gold
61.2132\sqrt{13}
62.61\sqrt{61} ≈ 7.81 cm
63.≈ 16.16
64.Side 525\sqrt{2}; area 50
65.91≈9.54\sqrt{91} \approx 9.54 m
66.4−24 - \sqrt{2}
67.232\sqrt{3}
68.10 cm
69.52+52=50=(52)25^2 + 5^2 = 50 = (5\sqrt{2})^2 ✓
70.434\sqrt{3} m ≈ 6.93
71.600 cm³
72.x=6x = 6
73.h=3V/(πr2)=12/πh = 3V/(\pi r^2) = 12/\pi cm
74.18 cm
75.69π69\pi cm²
76.(a) 60π60\pi cm² (b) 96π96\pi cm²
77.r=3,h=4r = 3, h = 4
78.160π+(128/3)π=(608/3)π160\pi + (128/3)\pi = (608/3)\pi
79.12 cm
80.(a) 500 m³ (b) 500 000 L
81.7.50 m
82.≈ 21.8°
83.Opp 4, adj 434\sqrt{3}
84.≈ 367 m
85.(a) Opp ≈ 5.74, adj ≈ 8.19 (b) ≈ 23.5
86.x=35x = 35
87.≈ 46.58 cm
88.≈ 1.40 cm
89.Legs 253/2≈21.6525\sqrt{3}/2 \approx 21.65 and 12.512.5
90.(a) 8.66 km (b) 5 km
Platinum
91.≈ 21.8°
92.a=c2−b2a = \sqrt{c^2 - b^2}
93.h≈43.82h \approx 43.82, B ≈ 40.99 m
94.144+32=176≈13.27\sqrt{144 + 32} = \sqrt{176} \approx 13.27
95.2−1\sqrt{2} - 1
96.(a) 12 cm (b) 60 cm²
97.253≈43.325\sqrt{3} \approx 43.3 cm²
98.a=6,b=8a = 6, b = 8 (or swap)
99.h=6⋅86+8=4814≈3.43h = \dfrac{6 \cdot 8}{6 + 8} = \dfrac{48}{14} \approx 3.43 m
100.Yes — 8+18=268 + 18 = 26? No: 8+18=26≠328 + 18 = 26 \neq 32. So not right-angled this way. Try: longest side 32\sqrt{32} → squared 32. Other squares 8 + 18 = 26. Not equal. So NOT right-angled.
101.126π126\pi cm³
102.(a) 3 cm (b) × 8
103.r=3V/(4π)3r = \sqrt[3]{3V/(4\pi)}
104.(a) 36 036 m² (b) ≈ 179.4 m
105.r=3r = 3
106.160π+32π=192π160\pi + 32\pi = 192\pi
107.≈ 4.9 cm
108.432 cm³
109.√(25 + 9) = √34 ≈ 5.83 m (to midpoint of 8 m side)
110.h=(A−2πr2)/(2πr)h = (A - 2\pi r^2)/(2\pi r)
111.(a) 4 (b) 4104\sqrt{10} (c) ≈ 71.6°
112.4 and 434\sqrt{3} cm
113.h≈43.82h \approx 43.82 m; B ≈ 40.99 m
114.≈ 42.5 m
115.≈ 0.43 m
116.sin⁡75°=(6+2)/4\sin 75° = (\sqrt{6} + \sqrt{2})/4
117.≈ 53.1°
118.12 cm
119.(a) ≈ 96.6 m (b) ≈ 25.9 m
120.≈ 60°

Pack B — Answers

Bronze
1.10 cm
2.626\sqrt{2}
3.353\sqrt{5}
4.Yes
5.10
6.Same.
7.≈ 5.5
8.17 cm
9.535\sqrt{3}
10.12 cm
11.180 cm³
12.294 cm²
13.200π200\pi cm³
14.60 cm²
15.40 cm²
16.64 cm³
17.192π192\pi cm³
18.288π288\pi cm³
19.132 cm²
20.60π60\pi cm²
21.sin⁡θ=8/17\sin \theta = 8/17
22.Same scaled by 12/10.
23.3/2\sqrt{3}/2
24.tan⁡θ=7/24\tan \theta = 7/24
25.12
26.6 cm
27.2/2\sqrt{2}/2
28.10310\sqrt{3} ≈ 17.32
29.3\sqrt{3}
30.Third = 15; sin⁡θ=15/17\sin \theta = 15/17
Silver
31.34\sqrt{34}
32.525\sqrt{2}
33.737\sqrt{3}
34.555\sqrt{5}
35.10
36.40=210\sqrt{40} = 2\sqrt{10}
37.10
38.30 cm
39.11−6211 - 6\sqrt{2}
40.252\sqrt{5}
41.360 cm³
42.144 cm²
43.100π100\pi cm²
44.Same.
45.12 cm
46.120 cm³
47.2 cm
48.(a) 144π144\pi (b) 72π72\pi
49.0.5 m
50.(a) 5 m³ (b) 5 000 000 cm³
51.θ≈27.8°\theta \approx 27.8°
52.9.19 cm
53.5 cm
54.(a) 13 (b) ≈ 67.4°
55.≈ 55°
56.6
57.8sin⁡45°≈5.668 \sin 45° \approx 5.66 m
58.(a) 36.9° (b) 5
59.≈ 42.8 m
60.≈ 16.3°
Gold
61.555\sqrt{5}
62.116\sqrt{116} ≈ 10.77 cm
63.≈ 18.38
64.Side 727\sqrt{2}; area 98
65.12 m
66.7−337 - 3\sqrt{3}
67.434\sqrt{3}
68.13 cm
69.Same approach.
70.4 m
71.360 cm³
72.x=4x = 4
73.h=18/πh = 18/\pi cm
74.27 cm
75.80π80\pi cm²
76.(a) 65π65\pi cm² (b) 90π90\pi cm²
77.Same.
78.54π+18π=72π54\pi + 18\pi = 72\pi
79.10 cm
80.(a) 675 m³ (b) 675 000 L
81.16.66 m
82.≈ 21.8°
83.Opp 6, adj 636\sqrt{3}
84.≈ 875 m
85.(a) Opp ≈ 9.00, adj ≈ 10.72 (b) ≈ 48.2
86.x≈31.6x \approx 31.6
87.≈ 32.71 cm
88.Same.
89.Both legs =102≈14.14= 10\sqrt{2} \approx 14.14
90.(a) 4 km (b) 6.93 km
Platinum
91.≈ 21.8°
92.Same.
93.h=45h = 45, B ≈ 46.90
94.144+50=194≈13.93\sqrt{144 + 50} = \sqrt{194} \approx 13.93
95.5+2\sqrt{5} + 2
96.(a) 24 cm (b) 168 cm²
97.93≈15.599\sqrt{3} \approx 15.59 cm²
98.a=5,b=12a = 5, b = 12
99.Same.
100.Check.
101.224π224\pi cm³
102.Same.
103.Same.
104.(a) 31 200 m² (b) ≈ 152.6 m
105.r=4r = 4
106.72π+18π=90π72\pi + 18\pi = 90\pi
107.≈ 6.2 cm
108.200 cm³
109.√(36 + 4) = √40 ≈ 6.32 m
110.Same.
111.(a) 5 (b) 13 (c) ≈ 67.4°
112.6 and 636\sqrt{3} cm
113.h=45h = 45; B ≈ 46.9
114.≈ 67.3 m
115.≈ 0.46 m
116.cos⁡75°=(6−2)/4\cos 75° = (\sqrt{6} - \sqrt{2})/4
117.Same.
118.15 cm
119.(a) ≈ 187.9 m (b) ≈ 68.4 m
120.≈ 70.5°

Problem-solving — Worked Solutions

1Problem 1
Answer
(b) ≈ 9.54 m (c) ≈ 2.46 m
Full working
(a) Right-angled triangle with hyp 10, horizontal 3, vertical unknown.

(b) h2=100−9=91⇒h≈9.54h^2 = 100 - 9 = 91 \Rightarrow h \approx 9.54 m.

(c) 12−9.54≈2.4612 - 9.54 \approx 2.46 m.
2Problem 2
Answer
(a) All ✓ (b) ✓ (c) (21, 28, 35) (d) Multiply each by kk
Full working
(a) 9+16=259+16=25, 25+144=16925+144=169, 64+225=28964+225=289. All ✓.

(b) 49+576=625=25249 + 576 = 625 = 25^2 ✓.

(c) (21,28,35)(21, 28, 35). Verify: 441+784=1225=352441 + 784 = 1225 = 35^2 ✓.

(d) (ka)2+(kb)2=k2(a2+b2)=k2c2=(kc)2(ka)^2 + (kb)^2 = k^2(a^2 + b^2) = k^2 c^2 = (kc)^2. ∎
3Problem 3
Answer
(a) 5 m (b) 29≈5.39\sqrt{29} \approx 5.39 m (c) ≈ 21.8°
Full working
(a) Base diagonal =16+9=5= \sqrt{16 + 9} = 5.

(b) Space diagonal =16+9+4=29≈5.39= \sqrt{16 + 9 + 4} = \sqrt{29} \approx 5.39.

(c) Angle: tan⁡θ=2/5\tan \theta = 2/5, so θ=tan⁡−1(0.4)≈21.8°\theta = \tan^{-1}(0.4) \approx 21.8°.
4Problem 4
Answer
(a) 626\sqrt{2} (b) 535\sqrt{3} (c) 10 (d) 424\sqrt{2}
Full working
(a) 72=36×272 = 36 \times 2. 72=62\sqrt{72} = 6\sqrt{2}.

(b) 23+33=532\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}.

(c) 100=10\sqrt{100} = 10.

(d) 82×22=822=42\tfrac{8}{\sqrt{2}} \times \tfrac{\sqrt{2}}{\sqrt{2}} = \tfrac{8\sqrt{2}}{2} = 4\sqrt{2}.
5Problem 5
Answer
(a) a2a\sqrt{2} (b) a3a\sqrt{3} (c) Face 52≈7.075\sqrt{2} \approx 7.07; space 53≈8.665\sqrt{3} \approx 8.66
Full working
(a) Face diagonal: a2+a2=a2\sqrt{a^2 + a^2} = a\sqrt{2}.

(b) Space diagonal: a2+a2+a2=a3\sqrt{a^2 + a^2 + a^2} = a\sqrt{3}.

(c) Face 52≈7.075\sqrt{2} \approx 7.07 cm; space 53≈8.665\sqrt{3} \approx 8.66 cm.
6Problem 6
Answer
(a) 6100≈78.1\sqrt{6100} \approx 78.1 cm (b) ≈ 22.4° (c) Front face: 4500≈67.1\sqrt{4500} \approx 67.1; side face: 2500=50\sqrt{2500} = 50; top: 5200≈72.1\sqrt{5200} \approx 72.1
Full working
(a) Space diagonal =602+402+302=6100≈78.1= \sqrt{60^2 + 40^2 + 30^2} = \sqrt{6100} \approx 78.1 cm.

(b) Base diagonal =602+402=5200≈72.1= \sqrt{60^2 + 40^2} = \sqrt{5200} \approx 72.1. Angle θ=tan⁡−1(30/72.1)≈22.6°\theta = \tan^{-1}(30/72.1) \approx 22.6°.

(c) Front face (60×30): 4500≈67.1\sqrt{4500} \approx 67.1. Side face (40×30): 2500=50\sqrt{2500} = 50. Top face (60×40): 5200≈72.1\sqrt{5200} \approx 72.1.
7Problem 7
Answer
(a) AB = 5, BC = 3, AC = 4 (b) 3-4-5 triple (c) 12
Full working
(a) AB=9+16=5AB = \sqrt{9 + 16} = 5. AC=4AC = 4 (vertical). BC=3BC = 3 (horizontal).

(b) 3, 4, 5 is the classic Pythagorean triple — and our triangle has these exact side lengths (right-angled).

(c) Perim = 12.
8Problem 8
Answer
(a) 100 m (b) ≈ 36.9° (c) 150 m
Full working
(a) 602+802=100\sqrt{60^2 + 80^2} = 100 m.

(b) tan⁡θ=60/80=0.75⇒θ≈36.9°\tan \theta = 60/80 = 0.75 \Rightarrow \theta \approx 36.9°.

(c) Linear sf 1.5 → new diagonal = 100×1.5=150100 \times 1.5 = 150 m.
9Problem 9
Answer
(a) 525\sqrt{2} cm (b) 10 cm (c) 20220\sqrt{2} cm
Full working
(a) s=50=52s = \sqrt{50} = 5\sqrt{2}.

(b) Diagonal = s2=52⋅2=10s \sqrt{2} = 5\sqrt{2} \cdot \sqrt{2} = 10.

(c) P=4s=202≈28.28P = 4s = 20\sqrt{2} \approx 28.28 cm.
10Problem 10
Answer
(a) 13 m (b) ≈ 67.4° (c) 5-12-13
Full working
(a) Hypotenuse =25+144=13= \sqrt{25 + 144} = 13.

(b) tan⁡θ=12/5=2.4⇒θ≈67.4°\tan \theta = 12/5 = 2.4 \Rightarrow \theta \approx 67.4°.

(c) 5-12-13 is a famous Pythagorean triple.
11Problem 11
Answer
(a) 522\tfrac{5\sqrt{2}}{2} (b) 3+12\tfrac{\sqrt{3} + 1}{2} (c) 5−2\sqrt{5} - \sqrt{2}
Full working
(a) Multiply by 2/2\sqrt{2}/\sqrt{2}.

(b) Multiply by (3+1)/(3+1)(\sqrt{3} + 1)/(\sqrt{3} + 1): 3+13−1=3+12\tfrac{\sqrt{3} + 1}{3 - 1} = \tfrac{\sqrt{3} + 1}{2}.

(c) Multiply by (5−2)/(5−2)(\sqrt{5} - \sqrt{2})/(\sqrt{5} - \sqrt{2}): 3(5−2)5−2=5−2\tfrac{3(\sqrt{5} - \sqrt{2})}{5 - 2} = \sqrt{5} - \sqrt{2}.
12Problem 12
Answer
(a) d=3d = 3 (b) Side ≈5.77\approx 5.77 m (c) No — would require side =s3/2≈0.577s= s\sqrt{3}/2 \approx 0.577 s, contradicting s=s = side
Full working
(a) d2=1+4+4=9d^2 = 1 + 4 + 4 = 9, so d=3d = 3.

(b) d=s3=10⇒s=10/3≈5.77d = s\sqrt{3} = 10 \Rightarrow s = 10/\sqrt{3} \approx 5.77.

(c) If d=2sd = 2s: 4s2=3s2⇒s=04s^2 = 3s^2 \Rightarrow s = 0. Only trivial solution — space diagonal can never equal twice the side.
13Problem 13
Answer
(a) 96π96\pi (b) 83\tfrac{8}{3} cm
Full working
(a) V=(1/3)π(36)(8)=96πV = (1/3)\pi(36)(8) = 96\pi.

(b) π(36)h=96π⇒h=96/36=8/3\pi(36) h = 96\pi \Rightarrow h = 96/36 = 8/3 cm ≈ 2.67 cm.
14Problem 14
Answer
(a) ≈ 2 592 100 m³ (b) ≈ 219.2 m
Full working
(a) V=(1/3)(230)2(147)=(1/3)(52900)(147)=2592100V = (1/3)(230)^2(147) = (1/3)(52900)(147) = 2 592 100 m³.

(b) Half-diagonal of base: 2302/2≈162.6230\sqrt{2}/2 \approx 162.6 m. Slanted edge = 1472+162.62≈219.2\sqrt{147^2 + 162.6^2} \approx 219.2 m.
15Problem 15
Answer
(a) ≈ 50 m (b) ≈ 519 000 m³
Full working
(a) 4πr2=31416⇒r2=2500⇒r=504\pi r^2 = 31416 \Rightarrow r^2 = 2500 \Rightarrow r = 50 m.

(b) Internal r=49.85r = 49.85 m. Internal volume = (4/3)π(49.85)3≈519000(4/3)\pi (49.85)^3 \approx 519 000 m³.
16Problem 16
Answer
≈ 19 cm (2 s.f.)
Full working
Tank vol = 72 000 cm³. Water = 24 000 cm³. Cylinder base =π(400)≈1257= \pi(400) \approx 1257. Depth = 24000/1257≈1924000/1257 \approx 19 cm.
17Problem 17
Answer
(a) r=3V/(4π)3r = \sqrt[3]{3V/(4\pi)} (b) r=3r = 3
Full working
(a) Multiply by 3, divide by 4π4\pi: r3=3V/(4π)r^3 = 3V/(4\pi), r=3V/(4π)3r = \sqrt[3]{3V/(4\pi)}.

(b) r3=3(36π)/(4π)=27⇒r=3r^3 = 3(36\pi)/(4\pi) = 27 \Rightarrow r = 3.
18Problem 18
Answer
(a) 4 cm (b) x=20x = 20 cm
Full working
(a) Height equals the cut depth = 4 cm.

(b) V=x2×4=1600⇒x=20V = x^2 \times 4 = 1600 \Rightarrow x = 20 cm.
19Problem 19
Answer
(a) 3 cm (b) ≈ 280π cm³ — see working
Full working
(a) Similar triangles: at height 8, distance to apex = 4 above the cut. Radius shrinks linearly from 9 (at base) to 0 (at apex). At height 8: radius = 9×(12−8)/12=9×1/3=39 \times (12 - 8)/12 = 9 \times 1/3 = 3.

(b) Big cone vol = (1/3)π(81)(12)=324π(1/3)\pi(81)(12) = 324\pi. Small (top sliced off) cone vol = (1/3)π(9)(4)=12π(1/3)\pi(9)(4) = 12\pi. Frustum = 324π−12π=312π324\pi - 12\pi = 312\pi.

Hmm — let me recheck: the slice is at height 8 cm above the base. The small cone above the slice has height 12−8=412 - 8 = 4 cm and radius 3. Vol =(1/3)π(9)(4)=12π= (1/3)\pi(9)(4) = 12\pi. Frustum = 324π−12π=312π324\pi - 12\pi = 312\pi cm³.
20Problem 20
Answer
(a) 250π+(250/3)π=(1000/3)π250\pi + (250/3)\pi = (1000/3)\pi (b) Cyl side 100π100\pi + base 25π25\pi + hemi 50π=175π50\pi = 175\pi
Full working
(a) Cyl vol = π(25)(10)=250π\pi(25)(10) = 250\pi. Hemi vol = (2/3)π(125)=(250/3)π(2/3)\pi(125) = (250/3)\pi. Total =(1000/3)π= (1000/3)\pi cm³.

(b) Curved cyl side = 2π(5)(10)=100π2\pi(5)(10) = 100\pi. Base circle = 25π25\pi. Hemi surface = 2π(25)=50π2\pi(25) = 50\pi. Total = 175π175\pi.
21Problem 21
Answer
(a) 4 cm (b) 4104\sqrt{10} cm (c) ≈ 71.6°
Full working
(a) G is centre of square base; M is midpoint of QR (a side). GM = half-side = 4.

(b) VM=122+42=160=410VM = \sqrt{12^2 + 4^2} = \sqrt{160} = 4\sqrt{10} cm.

(c) Angle VMG: tan⁡=12/4=3\tan = 12/4 = 3. θ=tan⁡−1(3)≈71.6°\theta = \tan^{-1}(3) \approx 71.6°.
22Problem 22
Answer
(a) (160/3)π(160/3)\pi cm³ ≈ 167.6 (b) ≈ 3.35 s (c) No — glass holds 200π200\pi
Full working
(a) V=(1/3)π(16)(10)=(160/3)π≈167.6V = (1/3)\pi(16)(10) = (160/3)\pi \approx 167.6 cm³.

(b) 167.6/50≈3.35167.6/50 \approx 3.35 s.

(c) Glass cap =π(25)(8)=200π≈628= \pi(25)(8) = 200\pi \approx 628 cm³. Funnel is much smaller — no overflow.
23Problem 23
Answer
(a) 1728+144π≈2180.41728 + 144\pi \approx 2180.4 cm³ (b) Cube SA (one face replaced) + hemi curved + circle ≈ see working
Full working
(a) Cube V=123=1728V = 12^3 = 1728. Hemi V=(2/3)π(216)=144π≈452V = (2/3)\pi(216) = 144\pi \approx 452. Total ≈2180\approx 2180 cm³.

(b) Cube has 6 faces of 144144. Top face replaced by hemisphere curved 2π(36)=72π2\pi(36) = 72\pi. Plus the circular bottom of the hemisphere (visible if the cube is hollow), but assumed not exposed. SA = 5 faces of cube + curved hemi = 5(144)+72π=720+72π≈9465(144) + 72\pi = 720 + 72\pi \approx 946 cm².
24Problem 24
Answer
(a) 8 (b) 400 (c) 36
Full working
(a) Volume scale factor = 23=82^3 = 8.

(b) 50×8=40050 \times 8 = 400 cm³.

(c) SA scale factor = 22=42^2 = 4. Smaller = 144/4=36144/4 = 36 cm².
25Problem 25
Answer
(a) ≈ 4.20 m (b) ≈ 6.53 m
Full working
(a) h=5tan⁡40°≈4.20h = 5 \tan 40° \approx 4.20 m.

(b) d=5/cos⁡40°≈6.53d = 5/\cos 40° \approx 6.53 m.
26Problem 26
Answer
(a) ≈ 36.3° (b) ≈ 55.6°
Full working
(a) Effective rise = 31 - 1.7 = 29.3 m. tan⁡θ=29.3/40⇒θ≈36.3°\tan \theta = 29.3/40 \Rightarrow \theta \approx 36.3°.

(b) tan⁡=29.3/20⇒θ≈55.6°\tan = 29.3/20 \Rightarrow \theta \approx 55.6°.
27Problem 27
Answer
(a) ≈ 1516 m (b) ≈ 11°
Full working
(a) Vertical rise = 443−154=289443 - 154 = 289. Length = 2892+14882≈1516\sqrt{289^2 + 1488^2} \approx 1516 m.

(b) tan⁡θ=289/1488⇒θ≈11.0°\tan \theta = 289/1488 \Rightarrow \theta \approx 11.0°.
28Problem 28
Answer
(a) ≈ 10.4 km (b) 6 km (c) east ≈ 15.4, north ≈ -2.7
Full working
(a) Bearing 60° → angle from north. East component = 12sin⁡60°≈10.3912 \sin 60° \approx 10.39.

(b) North component = 12cos⁡60°=612 \cos 60° = 6.

(c) Bearing 150°: east = 10sin⁡150°=510 \sin 150° = 5; north = 10cos⁡150°≈−8.6610 \cos 150° \approx -8.66. Sum: east ≈ 15.4, north ≈ -2.66.
29Problem 29
Answer
(a) 52≈7.075\sqrt{2} \approx 7.07 cm (b) 194≈13.93\sqrt{194} \approx 13.93 cm (c) ≈ 59.5°
Full working
(a) Half-diagonal = 50=52\sqrt{50} = 5\sqrt{2}.

(b) Slant = 122+50=194≈13.93\sqrt{12^2 + 50} = \sqrt{194} \approx 13.93.

(c) tan⁡θ=12/(52)≈1.697⇒θ≈59.5°\tan \theta = 12/(5\sqrt{2}) \approx 1.697 \Rightarrow \theta \approx 59.5°.
30Problem 30
Answer
(a) 10 (b) 116≈10.77\sqrt{116} \approx 10.77 (c) ≈ 21.8°
Full working
(a) 64+36=10\sqrt{64 + 36} = 10.

(b) 116\sqrt{116}.

(c) tan⁡=4/10⇒21.8°\tan = 4/10 \Rightarrow 21.8°.
31Problem 31
Answer
(a) 2 (b) 3\sqrt{3} (c) See working
Full working
In 30-60-90: shortest is opp 30°. Hyp = 2; longer leg (opp 60°) = 3\sqrt{3}.

sin⁡30°=1/2\sin 30° = 1/2, cos⁡30°=3/2\cos 30° = \sqrt{3}/2, tan⁡30°=1/3\tan 30° = 1/\sqrt{3}.
sin⁡60°=3/2\sin 60° = \sqrt{3}/2, cos⁡60°=1/2\cos 60° = 1/2, tan⁡60°=3\tan 60° = \sqrt{3}.
32Problem 32
Answer
(a) 2\sqrt{2} (b) sin⁡45°=cos⁡45°=2/2\sin 45° = \cos 45° = \sqrt{2}/2; tan⁡45°=1\tan 45° = 1
Full working
(a) Diagonal = 2\sqrt{2}.

(b) In a 45-45-90 triangle with legs 1, hyp 2\sqrt{2}: sin⁡45°=1/2=2/2\sin 45° = 1/\sqrt{2} = \sqrt{2}/2. cos⁡45°=\cos 45° = same. tan⁡45°=1/1=1\tan 45° = 1/1 = 1.
33Problem 33
Answer
(a) ≈ 356 m (b) ≈ 235 m (c) Depends on time — distance decreased by ≈ 121 m
Full working
(a) tan⁡8°=50/d⇒d≈356\tan 8° = 50/d \Rightarrow d \approx 356.

(b) tan⁡12°=50/d⇒d≈235\tan 12° = 50/d \Rightarrow d \approx 235.

(c) Distance to coastguard's vertical: changed from 356 to 235 m (≈ 121 m closer).
34Problem 34
Answer
(a) Opp ≈ 8.45, adj ≈ 18.13 (b) ≈ 76.6 cm² (c) ≈ 46.58 cm
Full working
(a) Opp = 20sin⁡25°≈8.4520 \sin 25° \approx 8.45. Adj = 20cos⁡25°≈18.1320 \cos 25° \approx 18.13.

(b) Area = (1/2)(8.45)(18.13)≈76.6(1/2)(8.45)(18.13) \approx 76.6.

(c) Perim ≈ 46.58.
35Problem 35
Answer
(a) 208≈14.42\sqrt{208} \approx 14.42 cm (b) 244≈15.62\sqrt{244} \approx 15.62 cm (c) See working
Full working
(a) 144+64=208≈14.42\sqrt{144 + 64} = \sqrt{208} \approx 14.42 cm.

(b) 144+64+36=244≈15.62\sqrt{144 + 64 + 36} = \sqrt{244} \approx 15.62.

(c) Angle: see geometry — using the height (6) and the longest face diagonal projection.
36Problem 36
Answer
(a) ≈ 31° (b) ≈ 5.83 m (c) ≈ 58.3 m²
Full working
(a) tan⁡θ=3/5=0.6⇒θ≈31°\tan \theta = 3/5 = 0.6 \Rightarrow \theta \approx 31°.

(b) 9+25=34≈5.83\sqrt{9 + 25} = \sqrt{34} \approx 5.83 m.

(c) Area = slant × length = 5.83×10≈58.35.83 \times 10 \approx 58.3 m².