Mathematics Standard • Year 11 • Module 2 • Lesson 15
Right-Angled Trig: Finding Angles — Past-Paper Style
HSC Mathematics Standard 2-style writing on inverse trig — short answers and one structured extended response with explicit marking criteria.
1. Short-answer questions
1.1 A right-angled triangle has a hypotenuse of 20 cm and one side of 12 cm.
(a) Find the angle between the 12 cm side and the hypotenuse, in degrees and minutes.
(b) State the other acute angle. 3 marks Band 3
1.2 A 10-metre ladder leans against a vertical wall. The foot of the ladder is 4.8 m from the base of the wall.
(a) Find the angle the ladder makes with the ground, in degrees and minutes.
(b) State the angle the ladder makes with the wall, in degrees and minutes. 3 marks Band 3-4
1.3 A pilot is flying at an altitude of 1500 m. She spots a runway 4200 m horizontally from directly below the aircraft.
(a) Sketch and label the right-angled triangle (right angle at the point directly below the aircraft).
(b) Find the angle of depression from the aircraft to the runway, in degrees and minutes. 4 marks Band 4
2. Extended response
2.1 A 15-metre telephone pole stands vertically. A safety regulation requires the support wire from the top of the pole to a ground anchor to make an angle of between 55° and 70° with the ground.
An installer first plans to anchor the wire 9 m from the base of the pole.
If the design fails the regulation, the installer must propose a new anchor distance that does satisfy it.
(a) For the proposed 9 m anchor, sketch and label the right-angled triangle, then find the angle between the wire and the ground in degrees and minutes.
(b) State whether the proposed design satisfies the regulation. Justify in one sentence.
(c) Find the range of anchor distances (in metres, to 1 d.p.) that would satisfy the regulation (i.e. the values for which the wire-to-ground angle is between 55° and 70°). State your answer with a clear conclusion sentence. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — labelled diagram with pole 15 m vertical and anchor 9 m horizontal.
• 1 mark — correct ratio tan θ = 15/9 with calculator step.
• 1 mark — correct angle in degrees and minutes.
Part (b) — 1 mark
• 1 mark — explicit yes/no with justification referencing the 55°-70° range and the computed angle.
Part (c) — 3 marks
• 1 mark — computes minimum anchor distance using θ = 70° (d_min = 15 ÷ tan 70°).
• 1 mark — computes maximum anchor distance using θ = 55° (d_max = 15 ÷ tan 55°).
• 1 mark — explicit conclusion sentence stating the range (e.g. "anchor distance must be between 5.5 m and 10.5 m").
Your response:
Stuck on (c)? Larger angle → smaller anchor distance, and vice versa. Min distance comes from θ = 70°; max distance comes from θ = 55°.How did this worksheet feel?
What I'll revisit before next class:
1.1 — H=20, side=12 (3 marks)
(a) Sample. The 12 cm side is adjacent to the angle (it shares the vertex with H). cos θ = 12/20 = 0.6 ⇒ θ = cos⁻¹(0.6) ≈ 53.13° = 53°8'.
(b) Sample. Other acute angle = 90° − 53°8' = 36°52'.
Marking notes. (a) 1 mark — chooses cos⁻¹ with substitution. 1 mark — answer in ° and '. (b) 1 mark — complementary subtraction in ° and '. Common error: using sin (would compute the OTHER acute angle, not the one asked).
1.2 — 10 m ladder, foot 4.8 m from wall (3 marks)
(a) Sample. cos⁻¹(4.8/10) = cos⁻¹(0.48) ≈ 61.31° = 61°19'.
(b) Sample. Angle with wall = 90° − 61°19' = 28°41'.
Marking notes. (a) 1 mark — correct ratio and substitution. 1 mark — answer in ° and ' (with the correct angle — the one at the ground, not at the wall). (b) 1 mark — uses complementary angle to get the wall angle. Common error: answering 61°19' for (b) — the wall angle is the complement.
1.3 — Pilot 1500 m altitude, runway 4200 m horizontal (4 marks)
(a) Sample diagram. Right angle at point directly below aircraft; vertical 1500 m down from aircraft to ground point; horizontal 4200 m from that ground point to runway; angle of depression marked at the aircraft between horizontal reference and line-of-sight to runway.
(b) Sample. tan θ = 1500/4200 = 0.3571; θ = tan⁻¹(0.3571) ≈ 19.65° = 19°39'.
Marking notes. (a) 1 mark — right angle at correct vertex. 1 mark — clear labels including the horizontal reference line at the aircraft (needed for depression). (b) 1 mark — correct ratio with substitution. 1 mark — correct answer in ° and '. Common error: measuring the angle from the vertical instead of from the horizontal (gives 70°21' instead of 19°39').
2.1 — Pole anchor regulation (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Diagram and angle for 9 m anchor.
Diagram: right angle at the base of the pole; vertical pole = 15 m; horizontal anchor distance = 9 m; wire = hypotenuse; angle θ between wire and ground (at the anchor). [1 mark — diagram.]
tan θ = 15 ÷ 9 = 1.6667. [1 mark — ratio.]
θ = tan⁻¹(1.6667) ≈ 59.04°. Decimal part: 0.04 × 60 ≈ 2'. So θ ≈ 59°2'. [1 mark — angle in ° and '.]
(b) Does the 9 m design satisfy the regulation?
55° ≤ 59°2' ≤ 70° → Yes, the proposed 9 m anchor satisfies the regulation. [1 mark — explicit yes with the inequality justification.]
(c) Range of allowed anchor distances.
Larger angle ⇒ smaller anchor distance. Use the regulation extremes (θ = 70° gives the minimum distance; θ = 55° gives the maximum distance).
Minimum: tan 70° = 15 ÷ d_min ⇒ d_min = 15 ÷ tan 70° ≈ 15 ÷ 2.7475 ≈ 5.5 m. [1 mark — min distance.]
Maximum: tan 55° = 15 ÷ d_max ⇒ d_max = 15 ÷ tan 55° ≈ 15 ÷ 1.4281 ≈ 10.5 m. [1 mark — max distance.]
Conclusion: the anchor distance must be between approximately 5.5 m and 10.5 m for the wire-to-ground angle to lie in the 55°-70° regulation range. The proposed 9 m anchor falls inside this range. [1 mark — explicit conclusion stating the range with units.]
Total: 7/7.
Band descriptors for marker.
Band 3: Diagram drawn and angle for 9 m anchor calculated; no part (c) attempted or only one bound found. ≈ 3-4 marks.
Band 4: Parts (a) and (b) correct; in (c), both bounds attempted but rounding errors or only one bound computed correctly. ≈ 5 marks.
Band 5: Full numerical solution with both bounds and conclusion, but conclusion missing units or omits the original 9 m comparison. ≈ 6 marks.
Band 6: Complete with diagram, both bounds correct, explicit conclusion sentence stating the allowed range in metres AND noting that 9 m falls within it. 7/7.