Mathematics • Year 9 • Unit 3 • Lesson 15
Two-Angle Trig — Mixed Challenge
Pull together everything from Unit 3 lessons 11–15: inverse trig, complementary angles, elevation, depression, the alternate-angles identity, and the two-equation method for unreachable heights. Choose the right tool per problem, spot a plausible Year 9 mistake, and tackle an open-ended challenge that designs a real-world two-angle scenario.
1. Mixed problems — choose the right rule
Each question uses a different combination of inverse trig, depression-equals-elevation, the two-equation method, and the complementary-angle rule. Decide which tool applies before you start writing. Show your working. 3 marks each
1.1 From a 60 m cliff, two boats are seen at depressions 35° and 15° (same straight line). (a) Find the gap between them (2 d.p.). (b) Verify the "smaller angle → further away" rule by stating which boat is further.
1.2 From point P on the ground, the angle of elevation to a flagpole top is 50°. From point Q, 15 m further from the pole on the same line, the elevation is 30°. Find the flagpole's height (2 d.p.).
1.3 An aircraft at altitude H flies horizontally over a flat field. It sights a marker on the ground at depression 12°. After flying 600 m in the direction of the marker (horizontally), the depression is now 28°. Find H (2 d.p.).
1.4 A flagpole sits on top of a school building. From a point on flat ground 50 m from the building's base, the angle of elevation to the BOTTOM of the flagpole (top of the building) is 30°, and to the TOP of the flagpole is 40°. Find the flagpole's length (2 d.p.).
1.5 A drone hovers at 40 m above flat ground. From the drone, two markers on the ground are seen at depressions 60° and 30°, BUT in OPPOSITE directions (one east, one west). Find the distance between the two markers (2 d.p.).
1.6 Two surveyors stand on flat ground 50 m apart, looking up at the SAME tower top. Surveyor A measures elevation 55°; surveyor B (on the opposite side of the tower) measures elevation 40°. The tower stands between A and B on the same straight line. Find the tower's height (2 d.p.) and its distance from each surveyor.
2. Find the mistake
A Year 9 student is asked: from a 75 m cliff, boat A is at depression 50° and boat B at depression 20° on the same line. Find the distance between them. Their working is shown. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — gap between two boats:
Line 1: d_A = 75 / tan 50° ≈ 75 / 1.1918 ≈ 62.92 m
Line 2: d_B = 75 / tan 20° ≈ 75 / 0.3640 ≈ 206.04 m
Line 3: Gap = d_A + d_B ≈ 62.92 + 206.04 ≈ 268.96 m
Line 4: The boats are about 268.96 m apart.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the correct gap.
Stuck? Both boats are on the SAME SIDE of the cliff (same straight line OUT from the base). So the gap between them is the SUBTRACTION of the closer distance from the farther distance — not the sum. (Sum would only apply if they were on opposite sides.)3. Open-ended challenge — design a two-angle scenario
This question has more than one valid answer. 4 marks
3.1 Design ONE realistic two-angle scenario that COMBINES elevation AND depression in the same problem (i.e. the observer looks both up and down). Then design ONE realistic two-angle scenario that uses two ELEVATIONS at DIFFERENT distances (the two-equation method to find an unreachable height). For EACH scenario:
(i) Describe the situation in 2–3 sentences.
(ii) State the two angles, the relevant heights or distances, and what is being asked.
(iii) Solve it.
(iv) Sense-check: are your answers realistic for what you described?
Bonus: Your two scenarios must be from DIFFERENT real-world contexts (e.g. one outdoor, one indoor; or one urban, one natural).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 60 m cliff, depressions 35° and 15°
(a) d_near = 60 / tan 35° ≈ 60 / 0.7002 ≈ 85.69 m. d_far = 60 / tan 15° ≈ 60 / 0.2679 ≈ 223.92 m. Gap = 223.92 − 85.69 ≈ 138.23 m.
(b) The 15° angle is smaller and corresponds to the further boat (223.92 m vs 85.69 m) — confirms "smaller depression = further away" ✓.
1.2 — Flagpole, elevations 50° and 30°, 15 m apart
Let h = pole height, d_P = distance from P (closer) to pole base. From P: h = d_P × tan 50°. From Q: h = (d_P + 15) × tan 30°.
Equate: d_P × tan 50° = (d_P + 15) × tan 30°.
d_P × (tan 50° − tan 30°) = 15 × tan 30°.
d_P = 15 × tan 30° / (tan 50° − tan 30°) ≈ 15 × 0.5774 / (1.1918 − 0.5774) ≈ 8.66 / 0.6144 ≈ 14.10 m.
h = 14.10 × tan 50° ≈ 14.10 × 1.1918 ≈ 16.80 m pole height.
1.3 — Aircraft, depressions 12° then 28° after 600 m closer
Let H = altitude, d = horizontal distance to the marker at the SECOND observation. From the second observation: H = d × tan 28°. From the first observation (600 m before, so 600 m further from marker): H = (d + 600) × tan 12°.
Equate: d × tan 28° = (d + 600) × tan 12°.
d × (tan 28° − tan 12°) = 600 × tan 12°.
d = 600 × tan 12° / (tan 28° − tan 12°) ≈ 600 × 0.2126 / (0.5317 − 0.2126) ≈ 127.56 / 0.3191 ≈ 399.74 m.
H = 399.74 × tan 28° ≈ 399.74 × 0.5317 ≈ 212.55 m altitude.
1.4 — Flagpole on a building, 50 m away, elevations 30° (bottom) and 40° (top)
Building height = 50 × tan 30° ≈ 50 × 0.5774 ≈ 28.87 m.
Top of flagpole height = 50 × tan 40° ≈ 50 × 0.8391 ≈ 41.95 m.
Flagpole length = 41.95 − 28.87 ≈ 13.08 m.
1.5 — Drone 40 m up, depressions 60° (east) and 30° (west)
d_east = 40 / tan 60° ≈ 40 / 1.7321 ≈ 23.09 m.
d_west = 40 / tan 30° ≈ 40 / 0.5774 ≈ 69.28 m.
Since the markers are on OPPOSITE SIDES of the drone, total distance between them = 23.09 + 69.28 ≈ 92.37 m.
1.6 — Two surveyors, 50 m apart, opposite sides of tower
Let d_A = distance from A to the tower base; then d_B = 50 − d_A (since tower is between A and B). h = d_A × tan 55° = (50 − d_A) × tan 40°.
d_A × tan 55° = (50 − d_A) × tan 40°.
d_A × (tan 55° + tan 40°) = 50 × tan 40°.
d_A = 50 × tan 40° / (tan 55° + tan 40°) ≈ 50 × 0.8391 / (1.4281 + 0.8391) ≈ 41.96 / 2.2672 ≈ 18.51 m.
Then d_B = 50 − 18.51 ≈ 31.49 m.
h = 18.51 × tan 55° ≈ 18.51 × 1.4281 ≈ 26.43 m tower height.
Sense check: h via B = 31.49 × tan 40° ≈ 31.49 × 0.8391 ≈ 26.42 m ✓ (matches within rounding).
2 — Find the mistake
(a) The mistake is on Line 3.
(b) Both boats are on the SAME side of the cliff (out from the cliff base, on the same straight line), so the gap between them is the SUBTRACTION of the closer distance from the further distance, NOT the sum. The sum would apply if they were on OPPOSITE sides of the cliff (which is impossible if both are out at sea from the same cliff base).
(c) Corrected working:
d_A = 75 / tan 50° ≈ 75 / 1.1918 ≈ 62.92 m (closer boat).
d_B = 75 / tan 20° ≈ 75 / 0.3640 ≈ 206.04 m (further boat).
Gap = d_B − d_A ≈ 206.04 − 62.92 ≈ 143.12 m.
Sense check: smaller depression (20°) → further boat (206 m); larger depression (50°) → closer boat (63 m). Gap = far − near. ✓
3 — Open-ended challenge (sample solution)
Scenario 1 (combined elevation + depression). "Sarah stands on a 6 m balcony in the city centre. She sees a window-washer rope dangling at angle of elevation 18° (above her) and a delivery van on the street below at angle of depression 35°. Both are at the same horizontal distance — 12 m — from her position. Find the vertical distance between the top of the rope (visible end) and the delivery van."
Solve: rope above eye = 12 × tan 18° ≈ 12 × 0.3249 ≈ 3.90 m. Van below eye = 12 × tan 35° ≈ 12 × 0.7002 ≈ 8.40 m. Total vertical separation ≈ 3.90 + 8.40 ≈ 12.30 m.
Sense check: realistic — the rope is hanging from a few floors above, and the van is on the street below, fitting a 4–5 storey building.
Scenario 2 (two-equation method). "A bushwalker tries to estimate the height of a mountain peak from two campsites in a valley. From campsite A, the peak's elevation is 18°. Walking 200 m further FROM the peak to campsite B, the elevation drops to 14°. Find the peak's height above the camp level."
Solve: Let h = peak height, d = horizontal distance from A to peak base. h = d × tan 18° (from A) and h = (d + 200) × tan 14° (from B, further away).
Equate: d × tan 18° = (d + 200) × tan 14°.
d × (tan 18° − tan 14°) = 200 × tan 14°.
d = 200 × tan 14° / (tan 18° − tan 14°) ≈ 200 × 0.2493 / (0.3249 − 0.2493) ≈ 49.86 / 0.0756 ≈ 659.52 m.
h = 659.52 × tan 18° ≈ 659.52 × 0.3249 ≈ 214.27 m above the camp level.
Sense check: a 214 m peak is realistic for a hill in eastern Australia at moderate distance.
Marking: 2 marks per scenario (1 for clear realistic setup, 1 for correct solve + sense check). Up to 4 in total. Combined elevation + depression scenario AND distinct two-equation scenario from a different context earns full marks; same-context pair earns 3/4.