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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 9 · 9.3 Right-angle Trigonometry

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.sin⁡θ=3/5\sin \theta = 3/5
2.Opp = 10sin⁡30°=510 \sin 30° = 5; adj = 10cos⁡30°=5310 \cos 30° = 5\sqrt{3}; hyp = 10
3.3/2\sqrt{3}/2
4.tan⁡θ=5/12\tan \theta = 5/12
5.6
6.4 cm
7.1
8.5
9.1/31/\sqrt{3} or 3/3\sqrt{3}/3
10.Third side = 5; sin⁡θ=5/13\sin \theta = 5/13
Silver
11.θ≈24.6°\theta \approx 24.6°
12.5.74 cm
13.4.62 cm
14.(a) 10 (b) ≈ 53.1°
15.≈ 32°
16.8.66
17.5sin⁡60°≈4.335 \sin 60° \approx 4.33 m
18.(a) 28.1° (b) 17
19.≈ 68.7 m
20.sin⁡−1(5/13)≈22.6°\sin^{-1}(5/13) \approx 22.6°
Gold
21.7.50 m
22.≈ 21.8°
23.Opp 4, adj 434\sqrt{3}
24.≈ 367 m
25.(a) Opp ≈ 5.74, adj ≈ 8.19 (b) ≈ 23.5
26.x=35x = 35
27.≈ 46.58 cm
28.≈ 1.40 cm
29.Legs 253/2≈21.6525\sqrt{3}/2 \approx 21.65 and 12.512.5
30.(a) 8.66 km (b) 5 km
Platinum
31.(a) 4 (b) 4104\sqrt{10} (c) ≈ 71.6°
32.4 and 434\sqrt{3} cm
33.h≈43.82h \approx 43.82 m; B ≈ 40.99 m
34.≈ 42.5 m
35.≈ 0.43 m
36.sin⁡75°=(6+2)/4\sin 75° = (\sqrt{6} + \sqrt{2})/4
37.≈ 53.1°
38.12 cm
39.(a) ≈ 96.6 m (b) ≈ 25.9 m
40.≈ 60°

Pack B — Answers

Bronze
1.sin⁡θ=8/17\sin \theta = 8/17
2.Same scaled by 12/10.
3.3/2\sqrt{3}/2
4.tan⁡θ=7/24\tan \theta = 7/24
5.12
6.6 cm
7.2/2\sqrt{2}/2
8.10310\sqrt{3} ≈ 17.32
9.3\sqrt{3}
10.Third = 15; sin⁡θ=15/17\sin \theta = 15/17
Silver
11.θ≈27.8°\theta \approx 27.8°
12.9.19 cm
13.5 cm
14.(a) 13 (b) ≈ 67.4°
15.≈ 55°
16.6
17.8sin⁡45°≈5.668 \sin 45° \approx 5.66 m
18.(a) 36.9° (b) 5
19.≈ 42.8 m
20.≈ 16.3°
Gold
21.16.66 m
22.≈ 21.8°
23.Opp 6, adj 636\sqrt{3}
24.≈ 875 m
25.(a) Opp ≈ 9.00, adj ≈ 10.72 (b) ≈ 48.2
26.x≈31.6x \approx 31.6
27.≈ 32.71 cm
28.Same.
29.Both legs =102≈14.14= 10\sqrt{2} \approx 14.14
30.(a) 4 km (b) 6.93 km
Platinum
31.(a) 5 (b) 13 (c) ≈ 67.4°
32.6 and 636\sqrt{3} cm
33.h=45h = 45; B ≈ 46.9
34.≈ 67.3 m
35.≈ 0.46 m
36.cos⁡75°=(6−2)/4\cos 75° = (\sqrt{6} - \sqrt{2})/4
37.Same.
38.15 cm
39.(a) ≈ 187.9 m (b) ≈ 68.4 m
40.≈ 70.5°

Problem-solving — Worked Solutions

1Problem 1
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.
2Problem 2
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°.
3Problem 3
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°.
4Problem 4
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.
5Problem 5
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°.
6Problem 6
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°.
7Problem 7
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}.
8Problem 8
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.
9Problem 9
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).
10Problem 10
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.
11Problem 11
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.
12Problem 12
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.