Mathematics Advanced • Year 11 • Module 2 • Lesson 4
Trigonometric Ratios
Practise HSC-style writing on trigonometric ratios, including the two-quadrant ambiguity and a multi-part extended-response problem.
1. Short-answer questions
1.1 Given sin θ = 3/5, find the two possible values of cos θ. State which quadrant each value corresponds to. 3 marks Band 4
1.2 A wheelchair ramp must have a gradient no steeper than 1 : 12 (rise : run). If the ramp makes an angle θ with the ground:
(a) Find tan θ.
(b) Hence find θ to the nearest degree. 3 marks Band 4
1.3 Given sin θ = −3/5 and tan θ = 3/4, determine the exact value of cos θ. State which quadrant θ lies in, with reasoning. 4 marks Band 5
Stuck on 1.3? Use the signs of sin and tan to pin down the quadrant, then use tan = sin/cos to solve for cos.2. Extended response
2.1 An observer at point O stands at the foot of a vertical cliff. A drone hovers above the cliff at point D. The horizontal distance from O to the base of the cliff (call it B) is 60 m. The angle of elevation from O to D is 50°, and the angle of elevation from O to the top of the cliff (point T) is 35°.
(a) Draw a clear right-triangle diagram showing O, B, T and D, and label every known angle and distance.
(b) Calculate the height of the cliff TB (i.e. from base to top), to 2 decimal places.
(c) Calculate the height of the drone above the cliff top (i.e. DT = DB − TB), to 2 decimal places.
(d) Without using a calculator, prove that the ratio (height of drone above ground) / (horizontal distance) equals tan 50°. Use this to verify your computed values are internally consistent. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• Clear diagram showing the three right triangles OBT and OBD sharing the side OB = 60 m, with labelled angles 35° at O looking up to T, 50° at O looking up to D, and 90° at B.
Part (b) — 2 marks
• 1 mark — sets up tan 35° = TB / 60 ⇒ TB = 60 tan 35°.
• 1 mark — computes TB ≈ 60 × 0.7002 ≈ 42.01 m.
Part (c) — 2 marks
• 1 mark — computes DB = 60 tan 50° ≈ 60 × 1.1918 ≈ 71.51 m.
• 1 mark — subtracts: DT = DB − TB = 71.51 − 42.01 = 29.50 m (allow ± 0.05 for rounding).
Part (d) — 3 marks
• 1 mark — identifies that in right triangle OBD, the angle at O is 50°; tan 50° = opp/adj = DB/OB.
• 1 mark — states DB = 60 tan 50° (matches (c)).
• 1 mark — verifies the computed DB / OB = 71.51 / 60 ≈ 1.1918 ≈ tan 50°. ✓
Your response:
Stuck on (d)? In triangle OBD, the height DB is opposite to the 50° angle at O.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Two possible cos θ from sin θ = 3/5 (3 marks)
Sample response. By the Pythagorean identity, cos²θ = 1 − (3/5)² = 1 − 9/25 = 16/25, so cos θ = ±4/5.
Since sin θ = 3/5 > 0, θ lies in QI or QII (where sin is positive).
• QI: cos θ = 4/5 (both cos and sin positive).
• QII: cos θ = −4/5 (sin positive, cos negative).
Marking notes. 1 mark — correct identity setup and cos²θ = 16/25. 1 mark — both values ±4/5 stated. 1 mark — labels each value with its quadrant (QI / QII). Common error: gives only the positive value (assumes acute) — loses 1 mark.
1.2 — Wheelchair ramp gradient 1:12 (3 marks)
Sample response.
(a) Gradient 1:12 means rise/run = 1/12, so tan θ = 1/12.
(b) θ = arctan(1/12) ≈ 4.76°, so to the nearest degree, θ = 5°.
Marking notes. 1 mark each: (a) correct tan ratio; (b) ≈ 4.76 with rounding shown; final 1 mark for the integer answer 5°.
1.3 — sin θ = −3/5, tan θ = 3/4 (4 marks)
Sample response. Sign analysis: sin < 0 (θ in QIII or QIV) and tan > 0 (θ in QI or QIII). The intersection is Quadrant III.
Now use tan θ = sin θ / cos θ: 3/4 = (−3/5) / cos θ, so cos θ = (−3/5) × 4/3 = −4/5.
Check via Pythagorean identity: sin²θ + cos²θ = 9/25 + 16/25 = 25/25 = 1 ✓.
Also check sign of cos: in QIII, cos is negative — consistent. ✓
Marking notes. 1 mark — identifies QIII via the sign-intersection of sin and tan. 1 mark — correctly sets up tan = sin/cos with substitution. 1 mark — arrives at cos θ = −4/5. 1 mark — verification (Pythagorean identity OR explicit ASTC sign check). Common error: just computes cos θ = +4/5 by dropping the sign halfway through, losing 1-2 marks.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Diagram. Three points marked: O (observer at ground), B (base of cliff, directly across from O, right angle at B), T (top of cliff above B), and D (drone above T). The line OB is horizontal, BT and TD are vertical. The angle TOB at the observer is 35°; the angle DOB at the observer is 50°. The horizontal distance OB = 60 m is labelled on the base. [1 mark]
Part (b) — Cliff height TB. In right triangle OBT, the angle at O is 35°, the adjacent side is OB = 60, and TB is opposite. So:
tan 35° = TB / 60 ⇒ TB = 60 tan 35° ≈ 60 × 0.7002 = 42.01 m. [2 marks]
Part (c) — Drone height above cliff top. In right triangle OBD, the angle at O is 50°:
tan 50° = DB / 60 ⇒ DB = 60 tan 50° ≈ 60 × 1.1918 = 71.51 m. [1 mark]
Drone above cliff top: DT = DB − TB = 71.51 − 42.01 = 29.50 m. [1 mark]
Part (d) — Internal consistency check. In right triangle OBD, the angle at O is 50° (by construction). The side opposite this angle is DB (the total height of drone above ground), and the side adjacent to it is OB = 60. By the definition of tangent:
tan 50° = opp / adj = DB / OB = DB / 60. [1 mark]
Rearranging: DB = 60 tan 50°. [1 mark] This matches the expression used in Part (c) — the diagram and the formula are internally consistent.
Numerical verification: DB / OB = 71.51 / 60 ≈ 1.1918 ≈ tan 50°. [1 mark] ✓ ▮
Total: 8/8.
Band descriptors for marker.
Band 3: Diagram partially correct (e.g. drone placed wrong); finds one of TB or DB correctly but not both; (d) attempted with poor reasoning. ≈ 3-4 marks.
Band 4: Diagram correct; (b) and (c) computed correctly; (d) gives the formula DB = 60 tan 50° but does not explicitly verify against the computed value. ≈ 5-6 marks.
Band 5: All parts complete; verification step in (d) makes explicit numerical comparison; minor presentation slip. ≈ 7 marks.
Band 6: Clear, fully labelled diagram; computation shown step-by-step; (d) derives the formula geometrically and validates it numerically; final answers given with appropriate units and rounding. 8/8. Top scripts also note that the question is geometrically well-posed precisely because the drone must lie above the cliff top — the angle of elevation to the drone must exceed that to the top (50° > 35° ✓).