Mathematics Standard • Year 11 • Module 2 • Lesson 14
Right-Angled Trig: Finding Sides — Past-Paper Style
HSC Mathematics Standard 2-style writing on SOHCAHTOA for unknown sides — short answers and one structured extended response with explicit marking criteria.
1. Short-answer questions
1.1 In a right-angled triangle, the angle θ = 34° and the hypotenuse is 20 cm. Calculate the side adjacent to the 34° angle, to 2 decimal places. Show the ratio used. 2 marks Band 3
1.2 A vertical pole has a wire attached from its top to the ground. The wire makes an angle of 65° with the ground and is anchored 4.5 m from the base of the pole.
(a) Sketch and label the right-angled triangle.
(b) Find the height of the pole, to 2 d.p. 3 marks Band 3-4
1.3 From a point on level ground, the angle of elevation to the top of a 60 m building is 41°.
(a) Find the distance from the point to the base of the building, to 1 d.p.
(b) The observer walks 30 m directly towards the building. Find the new horizontal distance to the base of the building, and state the new angle of elevation by re-doing the calculation, to the nearest minute. 4 marks Band 4
2. Extended response
2.1 A ship sails due east from port. After some time, the navigator observes a lighthouse on a bearing of N25°E (i.e. 25° east of north) from the ship. The lighthouse is known to be exactly 3.2 km due north of the port (so the lighthouse, the port and the ship form a right-angled triangle with the right angle at the ship's position).
The ship's fuel consumption is 9.0 L per nautical mile. (1 nautical mile ≈ 1.852 km.)
Marine diesel costs $2.40 per litre.
(a) Sketch and label the right-angled triangle formed by port, ship and lighthouse, marking the right angle at the ship and the 25° angle at the ship between north and the line to the lighthouse.
(b) Find the distance (in km, to 2 d.p.) the ship has sailed east of port, and the straight-line distance (in km, to 2 d.p.) from the ship to the lighthouse.
(c) Convert the ship's east-sailed distance into nautical miles (to 2 d.p.), then calculate the fuel cost for the eastbound leg to the nearest dollar. State your answer with a clear conclusion sentence. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — labelled right-angled triangle with the right angle at the ship, the 25° at the ship between N and the line to the lighthouse, north leg = 3.2 km, east leg unknown, hypotenuse = ship-to-lighthouse.
Part (b) — 3 marks
• 1 mark — chooses correct ratio: with the 25° at the ship and the north (3.2 km) side adjacent, the east leg is opposite — tan 25° = east ÷ 3.2.
• 1 mark — correct east distance = 3.2 × tan 25° to 2 d.p.
• 1 mark — correct ship-to-lighthouse distance H = 3.2 ÷ cos 25° to 2 d.p.
Part (c) — 3 marks
• 1 mark — correct unit conversion: east distance in km ÷ 1.852 → nautical miles.
• 1 mark — correct fuel cost = NM × 9.0 × $2.40, to the nearest dollar.
• 1 mark — explicit conclusion sentence with the dollar figure and units.
Your response:
Stuck on (c)? Convert km → NM first (÷1.852), then multiply by 9.0 L/NM and by $2.40/L.How did this worksheet feel?
What I'll revisit before next class:
1.1 — θ=34°, H=20 cm, find A (2 marks)
Sample. cos 34° = A ÷ 20 ⇒ A = 20 × cos 34° ≈ 20 × 0.8290 ≈ 16.58 cm.
Marking notes. 1 mark — chooses cos (A and H involved) with substitution shown. 1 mark — correct value to 2 d.p. with units. Common error: using sin (which gives the opposite side) — must label sides first.
1.2 — Pole with anchored wire (3 marks)
(a) Sample. Right angle at base of pole; 65° at the ground anchor; A = 4.5 m (horizontal anchor distance); O = pole height; H = wire.
(b) Sample. tan 65° = h ÷ 4.5 ⇒ h = 4.5 × tan 65° ≈ 4.5 × 2.1445 ≈ 9.65 m.
Marking notes. (a) 1 mark — labelled diagram with right angle at correct place. (b) 1 mark — correct ratio (tan) with substitution. 1 mark — correct value to 2 d.p. with units m. Common error: placing the right angle at the anchor instead of the base of the pole — produces the wrong ratio.
1.3 — 60 m building, observer walks closer (4 marks)
(a) Sample. tan 41° = 60 ÷ d ⇒ d = 60 ÷ tan 41° ≈ 60 ÷ 0.8693 ≈ 69.0 m.
(b) Sample. New distance = 69.0 − 30 = 39.0 m. New angle: tan θ' = 60 ÷ 39 ≈ 1.5385 ⇒ θ' = tan⁻¹(1.5385) ≈ 56.99° ≈ 56°59'.
Marking notes. (a) 1 mark — correct ratio + substitution. 1 mark — correct distance to 1 d.p. (b) 1 mark — correct new distance (39.0 m). 1 mark — correct new angle to the nearest minute. Common error: assuming the angle simply doubles when the distance halves.
2.1 — Ship-port-lighthouse (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Diagram. Right angle at the ship's position. North leg from ship up to lighthouse — length 3.2 km. East leg from ship back (west) to port — length = unknown east distance. Hypotenuse = ship-to-lighthouse. 25° angle at the ship between the north leg and the hypotenuse. [1 mark — diagram with right angle at ship, 25° at ship, 3.2 km labelled north.]
(b) East distance and ship-to-lighthouse distance.
North side is adjacent to the 25° angle (length 3.2 km). East side is opposite. Using tan: tan 25° = east ÷ 3.2. [1 mark — ratio.]
East = 3.2 × tan 25° ≈ 3.2 × 0.4663 ≈ 1.49 km. [1 mark — east distance to 2 d.p.]
Hypotenuse: cos 25° = 3.2 ÷ H ⇒ H = 3.2 ÷ cos 25° ≈ 3.2 ÷ 0.9063 ≈ 3.53 km. [1 mark — ship-to-lighthouse distance to 2 d.p.]
(c) Convert and cost.
East distance in NM = 1.49 ÷ 1.852 ≈ 0.80 NM. [1 mark — correct conversion to NM.]
Fuel cost = 0.80 × 9.0 L/NM × $2.40/L = 0.80 × $21.60 = $17.28 ≈ $17. [1 mark — multiplication and cost rounded to nearest dollar.]
Conclusion: the eastbound leg of about 1.49 km (≈0.80 nautical miles) costs about $17 in marine diesel. [1 mark — explicit conclusion sentence with dollar figure and units.]
Total: 7/7.
Band descriptors for marker.
Band 3: Diagram drawn with the right angle in the wrong location; finds one side using a ratio but mislabels O/A. ≈ 2 marks.
Band 4: Correct east distance and ship-to-lighthouse distance, but skips the NM conversion or computes cost in $/km instead of $/NM. ≈ 4 marks.
Band 5: Full numerical solution including the NM conversion, but conclusion sentence missing or numbers stated without units. ≈ 6 marks.
Band 6: Complete with correct ratios, correct conversions, and an explicit conclusion stating dollar cost with units. 7/7.