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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 11 · 11.7 Non-right-angle Trigonometry

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
Silver
11.11.7
12.48.3∘48.3^\circ
13.7.00
14.44.4∘44.4^\circ
15.(a) 52.2 m2^2; (b) BC≈13.0BC \approx 13.0 m
16.31.231.2 m
17.6.43 cm
18.North 25.0 km; East 43.3 km
19.636\sqrt{3}
20.≈299\approx 299 m
Gold
21.42.0∘42.0^\circ
22.∠PAB=90∘\angle PAB = 90^\circ; PB≈94.3PB \approx 94.3 km
23.≈87.3∘\approx 87.3^\circ
24.≈11.4\approx 11.4
25.≈45.6∘\approx 45.6^\circ
26.50=52\sqrt{50} = 5\sqrt{2} cm
27.≈25.1∘\approx 25.1^\circ
28.230∘230^\circ
29.B≈52.1∘B \approx 52.1^\circ or 127.9∘127.9^\circ
30.≈9.32\approx 9.32 km
Platinum
31.(a) 85∘85^\circ; (b) CL≈8.24CL \approx 8.24 km
32.≈32.0\approx 32.0 m
33.≈10.5\approx 10.5 km
34.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.
35.(a) 272\sqrt{7}; (b) 636\sqrt{3}
36.≈46.7∘\approx 46.7^\circ
37.≈49.3\approx 49.3 cm2^2
38.≈49.4\approx 49.4 km
39.≈232\approx 232 m
40.≈911\approx 911 km

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
Silver
11.10.3
12.62.1∘62.1^\circ
13.10.2
14.49.5∘49.5^\circ
15.(a) 74.8; (b) BC≈14.4BC \approx 14.4
16.37.337.3 m
17.15.3 cm
18.North −51.4-51.4 km (i.e. 51.4 km south); East 61.3 km
19.1534\dfrac{15\sqrt{3}}{4}
20.≈275\approx 275 m
Gold
21.75.4∘75.4^\circ
22.∠PAB=85∘\angle PAB = 85^\circ; PB≈77.5PB \approx 77.5 km
23.≈97.9∘\approx 97.9^\circ
24.≈12.0\approx 12.0
25.≈47.7∘\approx 47.7^\circ
26.77\sqrt{77} cm
27.≈27.1∘\approx 27.1^\circ
28.300∘300^\circ
29.B≈45.6∘B \approx 45.6^\circ or 134.4∘134.4^\circ
30.≈11.5\approx 11.5 km
Platinum
31.≈288∘\approx 288^\circ
32.≈30.1\approx 30.1 m
33.≈7.02\approx 7.02 km
34.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.
35.(a) 77; (b) 10310\sqrt{3}
36.≈48.5∘\approx 48.5^\circ
37.≈35.3\approx 35.3 cm2^2
38.≈49.0\approx 49.0 km
39.≈290\approx 290 m
40.≈645\approx 645 km

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.