Mathematics • Year 10 • Unit 3 • Lesson 20
Unit 3 Revision — Real-World Mixed Problems
Apply the full Unit 3 toolkit to authentic Australian scenarios — a cliff-top boat sighting, a yacht race map, a triangular nature reserve, a tangent-to-two-circles problem, and a cyclic quadrilateral on a survey plan. Pick the right tool at each step.
1. Word problems
For each: (i) draw a labelled diagram, (ii) identify the formula(s), (iii) calculate, (iv) round sensibly.
1.1 — Cliff-top boat sighting (drawn from Lesson 20 SA Q7). From a Sydney cliff 80 m above sea level, the angle of depression to a yacht is 25°. From the same cliff-top point, the angle of depression to a second yacht (further out, on the same bearing) is 40°.
(a) Find the distance from the foot of the cliff to each yacht.
(b) Find the distance between the two yachts.
(c) A third yacht is midway between them. What is its angle of depression from the cliff-top? 5 marks
1.2 — Yacht race triangle. A triangular race course has points A, B, C with AB = 15 km, ∠BAC = 50°, ∠ABC = 60°.
(a) Find sides AC and BC.
(b) Find the area enclosed by the course in km². 4 marks
1.3 — Nature reserve quadrilateral. A quadrilateral nature reserve ABCD has AB = 8 km, BC = 10 km, CD = 12 km, DA = 15 km, and diagonal AC = 14 km.
(a) Find ∠ABC using the cosine rule.
(b) Find ∠ADC using the cosine rule.
(c) Decide whether the reserve is a cyclic quadrilateral and justify. 5 marks
1.4 — Two circles, common tangent. Two circles of radii 5 cm and 12 cm touch externally. Find the length of their common external tangent. 3 marks
1.5 — Survey plan with similar triangles. On a survey plan, two similar triangles represent a scale drawing and the real paddock. The drawing triangle has sides 4, 5, 6 cm. The real paddock has perimeter 60 m.
(a) Find the scale factor (real ÷ drawing).
(b) Find the three real side lengths.
(c) Find the ratio of real area to drawing area. 3 marks
2. Explain your thinking
This question is about reasoning. Use full sentences. 4 marks
2.1 A classmate is preparing for the Unit 3 quiz and says: "I will just memorise every formula and apply them one by one." In 4-6 sentences, explain (i) why memorisation alone is not enough, (ii) what mental "decision tree" you actually use when you see a triangle problem (mention SAS, SSS, AAS), (iii) one piece of information that tells you to reach for a circle theorem rather than a triangle formula, and (iv) one type of error that pure memorisation will not catch but a sanity check will. Reference the words "included angle" and "cyclic" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cliff (80 m), depressions 25° and 40°
(a) d₁ = 80 / tan 25° = 80 / 0.4663 ≈ 171.6 m. d₂ = 80 / tan 40° = 80 / 0.8391 ≈ 95.3 m.
(b) Between yachts = 171.6 − 95.3 ≈ 76.3 m.
(c) Midway distance = (171.6 + 95.3)/2 ≈ 133.5 m. Angle of depression = tan⁻¹(80/133.5) = tan⁻¹(0.599) ≈ 30.9°.
1.2 — Race triangle (AB = 15, ∠A = 50°, ∠B = 60°)
(a) ∠C = 180° − 50° − 60° = 70°.
AC / sin 60° = 15 / sin 70° → AC = 15 × 0.866 / 0.9397 ≈ 13.8 km.
BC / sin 50° = 15 / sin 70° → BC = 15 × 0.766 / 0.9397 ≈ 12.2 km.
(b) Area = ½ × AB × AC × sin A = ½ × 15 × 13.8 × sin 50° = 103.5 × 0.766 ≈ 79.3 km².
1.3 — Quadrilateral reserve (AB = 8, BC = 10, CD = 12, DA = 15, AC = 14)
(a) In △ABC: cos B = (8² + 10² − 14²) / (2 × 8 × 10) = (64 + 100 − 196)/160 = −32/160 = −0.2. B = cos⁻¹(−0.2) ≈ 101.5°.
(b) In △ADC: cos D = (15² + 12² − 14²) / (2 × 15 × 12) = (225 + 144 − 196)/360 = 173/360 ≈ 0.4806. D = cos⁻¹(0.4806) ≈ 61.3°.
(c) B + D ≈ 101.5° + 61.3° = 162.8°, which is not 180°. So ABCD is not cyclic (opposite angles do not sum to 180°).
1.4 — Two circles, radii 5 and 12, external tangent
Distance between centres (externally touching) = 5 + 12 = 17. The external tangent forms a right-angled triangle with the line of centres (length 17) as hypotenuse and the difference of radii (12 − 5 = 7) as the perpendicular drop.
Tangent = √(17² − 7²) = √(289 − 49) = √240 = 4√15 ≈ 15.5 cm.
1.5 — Similar triangles, drawing 4-5-6 cm, real perimeter 60 m
(a) Drawing perimeter = 15 cm = 0.15 m. Scale factor = 60 / 0.15 = 400 (real is 400 × drawing).
(b) Real sides = 400 × 4 cm = 1 600 cm = 16 m; 400 × 5 cm = 20 m; 400 × 6 cm = 24 m. (Check: 16 + 20 + 24 = 60 m ✓.)
(c) Area ratio = 400² = 160 000 : 1.
2.1 — Explain your thinking (sample response)
Pure memorisation is not enough because every formula in Unit 3 is matched to a specific pattern of given information; without spotting the pattern, you cannot pick the right formula. My mental decision tree starts with "what am I given?": if I have two sides plus the included angle (SAS) I reach for the cosine rule (for the third side) or the area formula (for the area); if I have all three sides (SSS) I reach for the cosine rule (for an angle) or Heron's formula (for area); if I have two angles plus any one side (AAS / ASA) I reach for the sine rule. If the problem mentions a circle and four vertices on it, I switch to circle theorems instead of triangle rules — for example, opposite angles of a cyclic quadrilateral sum to 180°. Finally, memorisation will not catch sign errors (e.g. cos 120° is negative) or pattern errors (e.g. using the sine rule on pure SAS); a one-line sanity check — does the answer fit Pythagoras roughly? do the angles sum to 180°? — does.
Marking: 1 for naming the limitation of pure memorisation, 1 for an explicit decision tree referencing SAS/SSS/AAS and "included angle", 1 for naming a circle-theorem trigger (cyclic), 1 for the sanity-check argument.