Mathematics • Year 10 • Unit 3 • Lesson 20
Unit 3 — Master Mixed Challenge
The final push before Checkpoint 3. Mix every Unit 3 idea — right-trig, sine/cosine rules, area formulas, similarity, congruence, circle theorems and the SSA ambiguous case — then catch a classic exam-style error and design your own multi-step problem.
1. Mixed problems — name the rule, then solve
Before each calculation, write the rule or theorem you are using. Round to 1 dp / 0.1°. 2-3 marks each
1.1 Classify a triangle with sides 9, 40, 41 as acute / right / obtuse, and justify with one cosine-rule (or Pythagoras) calculation. 2 marks
1.2 Triangle has a = 11, b = 14, ∠C = 65°. Find c and the area. 3 marks
1.3 (Ambiguous case) Triangle with a = 8, b = 10, ∠A = 30°. How many distinct triangles fit? Find all possible values of ∠B. 3 marks
1.4 (Circle theorem) A tangent to a circle makes an angle of 50° with a chord at the point of contact. State the angle in the alternate segment and name the theorem. 2 marks
1.5 (Congruence) Two triangles have all three pairs of sides equal (SSS). Explain in one sentence why they must be congruent, and list the three other valid congruence tests from Unit 3. 2 marks
1.6 (Multi-step, Apollonius-style) Triangle ABC has AB = 15 km, ∠BAC = 50°, ∠ABC = 60°. (a) Find AC. (b) Find BC. (c) M is the midpoint of BC. Find AM using the cosine rule in △ABM. 3 marks
2. Find the mistake
A Year 10 student attempts the cyclic-quadrilateral check below (similar to Lesson 20 Short Answer Q6). They have AB = 8, BC = 10, CD = 12, DA = 15, AC = 14, and they correctly compute ∠ABC ≈ 101.5° and ∠ADC ≈ 61.3°. Their conclusion line is shown. Exactly one step contains a mistake. Spot it, explain why, then re-do correctly. 3 marks
Student's conclusion:
Line 1: In a cyclic quadrilateral, all four angles sum to 180°.
Line 2: ∠ABC + ∠ADC = 101.5° + 61.3° = 162.8°.
Line 3: Since 162.8° < 180°, the sum is not 180°.
Line 4: Therefore ABCD is cyclic.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong (and what the correct theorem says).
(c) Write the corrected conclusion (with the right "is / is not cyclic" verdict).
Stuck? There are two errors layered here. (i) The total of all four interior angles of any quadrilateral is 360°, not 180°. (ii) The cyclic-quadrilateral theorem is about opposite angles summing to 180°.3. Open-ended challenge — design your own Unit 3 question
Be creative but precise. Show every working. 4 marks
3.1 Design a multi-step Unit 3 question set in a realistic Australian context (paddock, bridge, ferry route, building, sports field …) that uses at least three different rules from Unit 3 — for example, the cosine rule for one side, the sine rule for one angle, and the area formula or Heron's for the area. You may optionally include one circle-theorem step.
For your question:
(i) State the context and all given numbers cleanly.
(ii) Draw a labelled diagram (with North if bearings are involved).
(iii) Solve all parts and report final answers with units, rounded sensibly.
Bonus: Verify one of your numerical answers using an alternative method (e.g. Heron's vs ½ ab sin C, or sum-of-angles = 180°). State the check explicitly.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Classify 9, 40, 41
9² + 40² = 81 + 1 600 = 1 681 = 41². Equal → right triangle. (Equivalently, cos C = (81 + 1600 − 1681)/(2·9·40) = 0 → C = 90°.) The 9-40-41 triple is a classic Pythagorean triple.
1.2 — a = 11, b = 14, C = 65°
c² = 121 + 196 − 308 cos 65° = 317 − 308 × 0.4226 = 317 − 130.2 = 186.8. c ≈ 13.7.
Area = ½ × 11 × 14 × sin 65° = 77 × 0.9063 ≈ 69.8.
1.3 — Ambiguous SSA, a = 8, b = 10, ∠A = 30°
Sine rule: sin B / 10 = sin 30° / 8 → sin B = 10 × 0.5 / 8 = 0.625.
B₁ = sin⁻¹(0.625) ≈ 38.7° (acute) — valid since 30° + 38.7° = 68.7° < 180°.
B₂ = 180° − 38.7° ≈ 141.3° (obtuse) — also valid since 30° + 141.3° = 171.3° < 180°.
Two distinct triangles fit the data (classic SSA ambiguous case).
1.4 — Tangent-chord angle 50°
Angle in alternate segment = 50°. Theorem: Alternate segment theorem — the angle between a tangent and a chord equals the inscribed angle in the alternate segment.
1.5 — SSS congruence
If all three pairs of sides are equal, the triangles have the same three side lengths and therefore the same shape and size — they must be congruent (SSS test). The other valid Unit 3 congruence tests are SAS, AAS, RHS.
1.6 — Apollonius-style problem
(a) ∠C = 180° − 50° − 60° = 70°. AC / sin 60° = 15 / sin 70° → AC ≈ 15 × 0.866 / 0.9397 ≈ 13.8 km.
(b) BC / sin 50° = 15 / sin 70° → BC ≈ 15 × 0.766 / 0.9397 ≈ 12.2 km.
(c) BM = 12.2 / 2 = 6.1 km. In △ABM: AM² = 15² + 6.1² − 2(15)(6.1) cos 60° = 225 + 37.21 − 91.5 = 170.71. AM = √170.71 ≈ 13.1 km.
2 — Find the mistake
(a) The mistake is on Line 1 (with a knock-on error in Line 4).
(b) The correct theorem: in a cyclic quadrilateral, opposite angles sum to 180° (not all four to 180°). The four interior angles of any quadrilateral sum to 360°. So the test is ∠ABC + ∠ADC = 180°, which here gives 162.8° ≠ 180°.
(c) Corrected conclusion: 162.8° ≠ 180°, so ABCD is not a cyclic quadrilateral. The student's verdict in Line 4 was the opposite of the correct one — even with the right calculation in Line 2, applying the wrong theorem in Line 1 led them to flip the answer.
3 — Open-ended (sample question and solution)
Scenario: A triangular section of bushland near Lithgow has corners A, B, C with AB = 20 km on bearing 075° from A, AC = 30 km with ∠BAC = 75°. (a) Find BC. (b) Find ∠ABC. (c) Find the area of the bushland in km².
(a) Cosine rule (SAS): BC² = 20² + 30² − 2(20)(30) cos 75° = 400 + 900 − 1200 × 0.2588 = 1300 − 310.6 = 989.4. BC = √989.4 ≈ 31.5 km.
(b) Sine rule: sin B / 30 = sin 75° / 31.5 → sin B = 30 × 0.9659 / 31.5 ≈ 0.920. B ≈ 66.9°.
(c) Area formula: A = ½ × 20 × 30 × sin 75° = 300 × 0.9659 ≈ 289.8 km².
Bonus check (Heron's): s = (20 + 30 + 31.5)/2 = 40.75. A = √(40.75 × 20.75 × 10.75 × 9.25) = √84 002 ≈ 289.8 km². ✓ Two methods agree.
Marking: 1 for a coherent scenario with clean numbers, 1 for using at least three different Unit 3 rules correctly, 1 for sensible rounding/units throughout, 1 for a valid alternative-method verification.