Mathematics • Year 9 • Unit 3 • Lesson 20
Trigonometry — Unit Mixed Challenge
The whole Year 9 trig kit on one page: Pythagoras, SOH-CAH-TOA, inverse trig, elevation/depression, compass and true bearings, components, multi-step. Spot a tool-selection mistake. Then design your own multi-tool trig adventure.
1. Mixed problems — pick the right tool
Each question targets a different combination of skills from across Unit 3. Sketch BEFORE you compute. Show your working. 3 marks each
1.1 A right triangle has hypotenuse 26 cm and one leg 24 cm. Find (a) the other leg (clean integer) and (b) the angles the hypotenuse makes with each leg (1 d.p.).
1.2 From a surveyor's station, the elevation to a hilltop is 18°. Moving 120 m closer, the elevation is 30°. Find the height of the hilltop above the surveyor's level (2 d.p.).
1.3 Convert the compass bearing S25°W to a true bearing, find its reverse, then express the reverse as a compass bearing.
1.4 A box is 5 cm by 12 cm by 84 cm. (a) Find the space diagonal (clean integer). (b) Find the angle the space diagonal makes with the base (1 d.p.).
1.5 A boat sails 24 km on true bearing 130°, then 18 km on true bearing 220°. Find (a) the straight-line distance from start and (b) the true bearing back to start (2 d.p. each).
1.6 From a 25 m cliff, two boats are seen on the same sight-line: boat A at angle of depression 40° and boat B at angle of depression 18°. Find the distance between the two boats (2 d.p.).
2. Find the mistake
A student has tried this mixed problem: "A 12 m flagpole casts a shadow 5 m long. Find the angle the sun's rays make with the ground." Their working is shown below. Exactly one line contains a mistake (a classic wrong-tool error). Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — 12 m flagpole, 5 m shadow, find sun-ray angle with ground:
Line 1: Right triangle: flagpole (vertical) = 12 m; shadow (horizontal) = 5 m; sun-ray = hypotenuse.
Line 2: Hyp = √(12² + 5²) = √169 = 13 m.
Line 3: Angle θ (with the ground) is opposite the flagpole side.
Line 4: sin θ = adj / hyp = 5 / 13 ≈ 0.3846.
Line 5: θ = sin−¹(0.3846) ≈ 22.6°.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer (1 d.p.).
Stuck? SOH-CAH-TOA: sin = opp/hyp, NOT adj/hyp. The student has used the wrong ratio name for the sides they actually picked.3. Open-ended challenge — design a multi-tool adventure
This question has many valid answers. 4 marks
3.1 Design a short word problem (3-5 sentences) that requires the solver to use AT LEAST THREE different tools from this unit. The pool of tools:
[T1] Pythagoras
[T2] sin/cos/tan to find a side
[T3] Inverse trig (sin−¹, cos−¹, tan−¹) to find an angle
[T4] Angle of elevation/depression
[T5] Compass bearings (form NθE etc.) or true bearings (three digits clockwise from N)
[T6] Reverse bearings (± 180°)
[T7] Component method (N = d cosθ, E = d sinθ) for multi-leg journeys
For your design, provide:
(a) The word problem itself.
(b) A list of which three+ tools are needed and where each one comes in.
(c) A full worked solution with intermediate answers labelled, rounded to a sensible precision at the end.
Bonus: mark in your worked solution where another student is most likely to make a mistake (e.g. mid-step rounding, wrong quadrant, wrong trig ratio) and what they should watch out for.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Hyp 26, leg 24
(a) Other leg = √(26² − 24²) = √(676 − 576) = √100 = 10 cm. (10-24-26 = 2×(5-12-13) — clean.)
(b) Angle opposite the 10: sin−¹(10/26) ≈ 22.6°. The other acute angle = 90 − 22.6 = 67.4°. The hypotenuse meets each leg at one of these two angles.
1.2 — Surveyor hilltop (18° from A, 30° from B, AB = 120 m)
From B: h = d tan 30°.
From A: h = (d + 120) tan 18°.
Equate: d tan 30° = (d + 120) tan 18° ⇒ d(tan 30° − tan 18°) = 120 tan 18°.
d = 120 × tan 18° / (tan 30° − tan 18°) ≈ 120 × 0.3249 / (0.5774 − 0.3249) ≈ 38.99 / 0.2525 ≈ 154.41 m.
h = 154.41 × tan 30° ≈ 154.41 × 0.5774 ≈ 89.16 m.
1.3 — S25°W → true → reverse → compass
S25°W in SW quadrant: true = 180 + 25 = 205°.
Reverse: 205 − 180 = 025°.
025° in NE quadrant: compass = N25°E.
Sanity check: S25°W and N25°E are exact opposites — a reverse pair, as expected.
1.4 — Box 5 by 12 by 84
(a) Space diagonal = √(25 + 144 + 7056) = √7225 = 85 cm (clean integer — uses 5-12-13 base → √(13² + 84²) = √(169 + 7056) = √7225 = 85; or 13-84-85 triple).
(b) Base diagonal = √(25 + 144) = 13. Angle with base: tan θ = 84 / 13 ≈ 6.4615; θ = tan−¹(6.4615) ≈ 81.2°.
1.5 — 24 km on 130 then 18 km on 220
Leg 1: N1 = 24 cos 130 ≈ −15.43; E1 = 24 sin 130 ≈ 18.39.
Leg 2: N2 = 18 cos 220 ≈ −13.79; E2 = 18 sin 220 ≈ −11.57.
Net N ≈ −29.22 (south); Net E ≈ 6.82 (east).
(a) Distance from start = √(29.22² + 6.82²) ≈ √(853.8 + 46.5) ≈ 30.01 km.
(b) End in SE quadrant. Angle from S = tan−¹(6.82/29.22) ≈ 13.13°. True bearing start → end = 180 − 13.13 ≈ 166.87°. Return bearing back to start = 166.87 + 180 = 346.87°.
1.6 — Two boats from a 25 m cliff (40° and 18° depression)
dA = 25 / tan 40° ≈ 25 / 0.8391 ≈ 29.79 m.
dB = 25 / tan 18° ≈ 25 / 0.3249 ≈ 76.95 m.
Distance between boats = dB − dA ≈ 76.95 − 29.79 ≈ 47.16 m.
2 — Find the mistake (12 m flagpole, 5 m shadow)
(a) The mistake is on Line 4.
(b) The student has used sin θ = adj/hyp, but SOH says sin = opp/hyp. The 5 m shadow is ADJACENT to the angle θ (at the base of the flagpole where the rays meet the ground), and the 12 m flagpole is OPPOSITE. So either: sin θ = 12/13 (using opp/hyp), OR cos θ = 5/13 (using adj/hyp), OR tan θ = 12/5 (using opp/adj). Any of those gives the correct angle; sin θ = 5/13 mixes up the rule.
(c) Corrected working:
Line 4 (option A): sin θ = opp/hyp = 12/13 ≈ 0.9231 ⇒ θ = sin−¹(0.9231) ≈ 67.4°.
Line 4 (option B): tan θ = opp/adj = 12/5 = 2.4 ⇒ θ = tan−¹(2.4) ≈ 67.4°. ✓
Sanity check: a 12 m flagpole casting only a 5 m shadow means the sun is HIGH in the sky — close to overhead — so the angle should be large (closer to 90°), not 22.6°. ✓
3 — Design a multi-tool adventure (sample solution)
(a) Word problem: "A bushwalker leaves base camp and walks 6 km on true bearing 040°, then 4 km on true bearing 130°, where she stops to camp. (i) Find her position relative to base camp (km N and km E, 2 d.p.). (ii) Find the straight-line distance from base camp to the new campsite. (iii) Find the true bearing she should walk to return DIRECTLY to base camp."
(b) Tools used: [T5/T7] bearings + components for each leg; [T1] Pythagoras for the straight-line distance; [T3] inverse trig (tan−¹) for the resultant bearing angle; [T6] reverse bearing (± 180°) for the return trip. That's 4 tools.
(c) Worked solution:
Leg 1: N1 = 6 cos 40 ≈ 4.60; E1 = 6 sin 40 ≈ 3.86.
Leg 2: N2 = 4 cos 130 ≈ −2.57; E2 = 4 sin 130 ≈ 3.06.
(i) Net N ≈ 2.03 km north; Net E ≈ 6.92 km east.
(ii) Distance = √(2.03² + 6.92²) ≈ 7.21 km.
(iii) NE quadrant; angle from N = tan−¹(6.92/2.03) ≈ 73.65°. True bearing camp → site ≈ 073.7°. Return = 073.7 + 180 ≈ 253.7°.
Bonus — likely mistakes: a classmate might (1) forget that cos 130° is NEGATIVE (south-ward N-component) and end up with a wrong net N; (2) get the right resultant bearing 074° but forget to add 180° for the RETURN trip; or (3) round 4.60 and 3.86 in step 1 then carry those rounded values forward to step (ii), losing precision.
Marking: 1 mark for a coherent word problem requiring 3+ tools; 1 mark for correctly listing which tools are needed where; 1 mark for a complete worked solution with sensible final rounding; 1 mark for the bonus identifying a likely classmate mistake.