Mathematics Standard • Year 11 • Module 2 • Lesson 16
Angles of Elevation and Depression — Past-Paper Style
HSC Mathematics Standard 2-style writing on elevation/depression — short answers and one structured two-observation extended response with explicit marking criteria.
1. Short-answer questions
1.1 From a point on level ground, the angle of elevation to the top of a vertical cliff is 34°. The point is 220 m from the base of the cliff.
(a) Draw a fully labelled diagram (horizontal reference line, angle, right angle at base, all known measurements).
(b) Find the height of the cliff, to 2 decimal places. 3 marks Band 3
1.2 From the top of a 50 m building, the angle of depression to a car on the street is 32°.
(a) Sketch and label the right-angled triangle. Mark the horizontal reference at the top and the equal alternate angle at the car.
(b) Find the horizontal distance from the base of the building to the car, to 2 decimal places. 3 marks Band 3-4
1.3 A 12 m antenna sits on top of a 40 m building. An observer stands 75 m from the base on level ground.
(a) Find the angle of elevation to the top of the building only, to the nearest minute.
(b) Find the angle of elevation to the top of the antenna, to the nearest minute.
(c) State the angle subtended by the antenna at the observer (the difference of (a) and (b)). 4 marks Band 4
2. Extended response (two-observation problem)
2.1 From point A on flat ground, the angle of elevation to the top of a vertical tower is 55°. From point B, which is 40 m further from the tower than A and on the same side (also on level ground), the angle of elevation is 35°.
Standard exam values: tan 55° ≈ 1.4281; tan 35° ≈ 0.7002.
(a) Sketch and label the diagram: tower vertical with right angle at base, points A (closer) and B (40 m further from tower than A) on level ground, with elevation angles 55° at A and 35° at B.
(b) Let d = horizontal distance from A to the tower base and h = tower height. Write two equations for h in terms of d.
(c) By setting the two equations equal, find d correct to 2 decimal places.
(d) Hence find the tower height h, to 2 decimal places, with a clear conclusion sentence. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — labelled diagram with A closer to the tower (larger angle 55°), B 40 m further away (smaller angle 35°), right angle at tower base.
Part (b) — 1 mark
• 1 mark — both equations correct: h = d × tan 55° AND h = (d + 40) × tan 35°.
Part (c) — 3 marks
• 1 mark — sets equations equal AND expands the right side.
• 1 mark — collects d-terms on one side, isolates d.
• 1 mark — correct numerical d ≈ 38.48 m.
Part (d) — 2 marks
• 1 mark — substitutes d back into either equation to get h ≈ 54.95 m.
• 1 mark — explicit conclusion sentence stating the tower height to 2 d.p. with units.
Your response:
Stuck on (c)? d × 1.4281 = (d + 40) × 0.7002. Expand: 1.4281d = 0.7002d + 28.008. Subtract 0.7002d both sides: 0.7279d = 28.008. Divide: d ≈ 38.48.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cliff elevation from 220 m (3 marks)
(a) Diagram. Horizontal ground line; observer at one end with horizontal reference; 34° angle of elevation marked from the horizontal up to the line of sight; vertical cliff at the other end with right angle at base; 220 m horizontal labelled; cliff height x labelled.
(b) Sample. tan 34° = x ÷ 220 ⇒ x = 220 × tan 34° ≈ 220 × 0.6745 ≈ 148.39 m.
Marking notes. 1 mark — diagram with horizontal reference and right angle at base. 1 mark — correct ratio + substitution. 1 mark — correct height to 2 d.p. with units. Common error: placing the angle between the cliff and the line-of-sight (which gives the complement instead).
1.2 — Depression from 50 m building (3 marks)
(a) Diagram. Building vertical (50 m), right angle at base; observer at top with horizontal reference; depression 32° below horizontal at top; equal-alternate elevation 32° at the car on the street.
(b) Sample. Elevation at car = 32° (alternate angles). tan 32° = 50 ÷ d ⇒ d = 50 ÷ tan 32° ≈ 50 ÷ 0.6249 ≈ 80.02 m.
Marking notes. (a) 1 mark — labelled diagram including the horizontal reference at the top. (b) 1 mark — correct equal-angle reasoning. 1 mark — correct value to 2 d.p. with units. Common error: using sin or cos instead of tan, or treating the depression angle as the wall-side angle.
1.3 — Antenna + building, observer 75 m (4 marks)
(a) Sample. tan⁻¹(40/75) = tan⁻¹(0.5333) ≈ 28.07° = 28°4'.
(b) Sample. Total height = 40 + 12 = 52 m. tan⁻¹(52/75) = tan⁻¹(0.6933) ≈ 34.72° = 34°43'.
(c) Sample. Subtended angle = 34°43' − 28°4' = 6°39'.
Marking notes. (a) 1 mark — correct angle in ° and '. (b) 1 mark — uses total height 52. 1 mark — correct angle in ° and '. (c) 1 mark — correct subtraction in ° and '. Common error in (c): reducing 43' − 4' = 39' but writing 7° instead of 6° (or vice versa) due to careless degree subtraction.
2.1 — Two-observation tower (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Diagram. Level ground line; tower vertical at one end with right angle at base; A on the ground closer to the tower (angle 55° at A from horizontal to top of tower); B on the ground, 40 m further from the tower than A (angle 35° at B from horizontal to top). [1 mark — labelled diagram.]
(b) Two equations.
From A: tan 55° = h ÷ d ⇒ h = d × tan 55°.
From B: tan 35° = h ÷ (d + 40) ⇒ h = (d + 40) × tan 35°. [1 mark — both equations.]
(c) Solve for d.
Setting equal: d × tan 55° = (d + 40) × tan 35°.
Expand: 1.4281d = 0.7002d + 28.008. [1 mark — set equal and expanded.]
Collect d: 1.4281d − 0.7002d = 28.008 ⇒ 0.7279d = 28.008. [1 mark — collected d and isolated.]
Divide: d = 28.008 ÷ 0.7279 ≈ 38.48 m. [1 mark — correct d to 2 d.p.]
(d) Find h.
h = 38.48 × tan 55° ≈ 38.48 × 1.4281 ≈ 54.95 m. [1 mark — back-substitution correct.]
Conclusion: the tower is approximately 54.95 m tall, with A located about 38.48 m from the base. [1 mark — explicit conclusion sentence with units.]
Total: 7/7.
Band descriptors for marker.
Band 3: Diagram drawn (perhaps with A and B reversed — larger angle at the farther point); only one equation written; no algebra. ≈ 2 marks.
Band 4: Both equations correct; sets them equal but algebraic manipulation has one error (e.g. forgets to expand or collect terms); d found incorrectly. ≈ 4 marks.
Band 5: Full numerical solution with both d and h correct, but conclusion missing or stated without units. ≈ 6 marks.
Band 6: Complete, correct, clearly laid out with diagram, both equations, full algebra and a final conclusion sentence stating h to 2 d.p. with units. 7/7.