Mathematics Advanced • Year 11 • Module 2 • Lesson 4
Trigonometric Ratios
Apply trig ratios to real engineering and surveying problems: ladder safety, accessibility ramps, surveying offset, roof pitch, and flagpole height.
Problem 1 — Ladder safety (the lesson's hook)
A 4-metre ladder leans against a wall, making an angle of 60° with the ground.
Set up: What are we solving for?
(i) Find the exact height (in metres) that the ladder reaches up the wall, using sin 60° = √3/2. Give a decimal approximation to 1 d.p. 2 marks
(ii) Find the horizontal distance from the wall to the base of the ladder (exact form and decimal to 1 d.p.). 2 marks
(iii) WorkSafe NSW recommends ladder angles between 70° and 80° for safety. The current 60° angle is too shallow. If the wall reach must remain 2√3 m (your answer from (i)), find the required ladder length if the angle is increased to 75°. (Use sin 75° ≈ 0.966.) 3 marks
Stuck on (iii)? Use sin 75° = (2√3) / L ⇒ L = (2√3) / sin 75°.Problem 2 — Wheelchair-ramp gradient (accessibility)
Australian building code AS 1428.1 requires wheelchair ramps to have a gradient no steeper than 1 : 14 (rise : run). The architect designs a ramp with horizontal run 7 m and rise 0.5 m.
Set up: What are we solving for?
(i) Find tan θ (where θ is the angle the ramp makes with the horizontal) and hence find θ to the nearest 0.1°. 2 marks
(ii) Does this ramp meet AS 1428.1? Compare the actual gradient with the 1:14 limit and state whether it passes. 2 marks
(iii) Find the length of the ramp itself (the hypotenuse, i.e. the slope length) to 2 decimal places. 2 marks
Problem 3 — Surveying triangle (data)
A surveyor wants to measure the distance across a river. From point A on one bank, she sights a tree T on the opposite bank directly across, then walks 50 m along her bank to point B and measures the angle ABT = 35° (the angle at B looking back to the tree).
Set up: What are we solving for?
(i) Sketch the triangle ABT (right-angled at A, since AT is "directly across"). Label A, B, T, and the side AT (unknown river width). 1 mark
(ii) Which trig ratio links the 35° angle, the 50 m walked distance (AB), and the river width AT? Set up the equation and solve for AT to 2 decimal places. (Use tan 35° ≈ 0.7002.) 3 marks
(iii) The surveyor also wants the direct distance BT (the sighting line). Find BT to 2 decimal places. (Use cos 35° ≈ 0.8192.) 2 marks
Stuck on (ii)? At B, AT is opposite the 35° angle and AB is adjacent — use tan.Problem 4 — Roof pitch from rise and run (modelling)
A house roof has a horizontal span of 8 m and a vertical rise of 3 m from eaves to ridge. The "pitch" is the angle between the rafter and the horizontal.
Set up: What are we solving for?
(i) Find tan(pitch) for the half-roof (using half the span = 4 m as adjacent and rise = 3 m as opposite). 2 marks
(ii) Find the pitch angle to the nearest 0.1°. 2 marks
(iii) The carpenter needs to know the rafter length (slope length from eaves to ridge). Find this in exact form using Pythagoras, then to 2 decimal places. 2 marks
Problem 5 — Flagpole shadow (real measurement)
A flagpole casts a shadow on level ground. At 9 am, the sun's altitude is 30° above the horizon and the shadow is 17.3 m long. At 12 noon, the sun's altitude is 60° and the shadow has shortened.
Set up: What are we solving for?
(i) Find the height of the flagpole (exact form). Use tan 30° = 1/√3 = √3/3 and treat the shadow length as the adjacent side. 2 marks
(ii) Find the length of the noon shadow (exact form, using tan 60° = √3). 2 marks
(iii) A student claims "the shadow at noon is one-third the morning shadow because tan 60° / tan 30° = 3". Evaluate the claim — is the ratio of shadows really 1:3, and is the student's reasoning correct? 2 marks
Stuck on (iii)? Compare shadownoon/shadowam; it should equal tan 30° / tan 60° (height cancels in both ratios).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Ladder safety
Set up. Right triangle with hypotenuse = ladder length and angle at the ground = 60°. We use sin (opp/hyp) for the wall reach, cos (adj/hyp) for the horizontal distance.
(i) sin 60° = h/4 ⇒ h = 4 × √3/2 = 2√3 m ≈ 3.5 m.
(ii) cos 60° = d/4 ⇒ d = 4 × 1/2 = 2 m.
(iii) sin 75° = (2√3)/L ⇒ L = 2√3 / sin 75° ≈ 2√3 / 0.966 ≈ 3.464 / 0.966 ≈ 3.59 m. (So a slightly shorter ladder — the steeper angle means the ladder doesn't need to extend as far up.)
Problem 2 — Wheelchair ramp
Set up. A right triangle with rise = 0.5 m (opposite), run = 7 m (adjacent), and slope = ramp length (hypotenuse).
(i) tan θ = 0.5/7 = 1/14. θ = arctan(1/14) ≈ 4.1°.
(ii) The gradient 1:14 is exactly the AS 1428.1 maximum. So the ramp meets the standard (it is on the boundary but not steeper). Passes.
(iii) Ramp length L = √(0.5² + 7²) = √(0.25 + 49) = √49.25 ≈ 7.02 m. (Very close to the run, because the angle is shallow.)
Problem 3 — Surveying triangle
Set up. Right triangle ABT, right-angled at A (since AT crosses directly to the tree). At B, angle = 35°; AB = 50 m (adjacent to 35°); AT = river width (opposite to 35°); BT = hypotenuse (sighting line).
(i) Sketch: A at one corner with right angle; B 50 m along the bank to one side; T directly across the river from A. AT is the river width.
(ii) tan 35° = opp/adj = AT/50 ⇒ AT = 50 × tan 35° ≈ 50 × 0.7002 = 35.01 m.
(iii) cos 35° = adj/hyp = 50/BT ⇒ BT = 50/cos 35° ≈ 50/0.8192 ≈ 61.04 m.
Problem 4 — Roof pitch
Set up. Half-roof right triangle: adj = 4 m (half the 8 m span), opp = 3 m (rise), hyp = rafter length.
(i) tan(pitch) = opp/adj = 3/4 = 0.75.
(ii) pitch = arctan(3/4) ≈ 36.9°. (A steep roof; common in alpine areas.)
(iii) Rafter = √(3² + 4²) = √25 = 5 m exactly (the classic 3-4-5 Pythagorean triple). To 2 d.p.: 5.00 m.
Problem 5 — Flagpole shadow
Set up. Right triangle: pole = opposite side, shadow = adjacent side, sun's altitude = angle at the ground.
(i) tan 30° = h/17.3 ⇒ h = 17.3 × tan 30° = 17.3 × √3/3 = 17.3√3/3. Numerically: 17.3 × 0.5774 ≈ 9.99 ≈ 10 m (since the given 17.3 was clearly chosen as 10√3). Exact form: h = 10 m.
(ii) tan 60° = 10/snoon ⇒ snoon = 10/√3 = 10√3/3 m ≈ 5.77 m. Exact form: 10√3/3 m.
(iii) The student's answer is wrong, but the structure is partly right. Actual ratio: snoon/sam = (10√3/3) / (10√3) = 1/3. So the noon shadow is one-third the morning shadow — the claim is correct. The student's reasoning "because tan 60° / tan 30° = 3" is also correct: s = h / tan(altitude), so snoon/sam = tan 30° / tan 60° = (√3/3) / √3 = 1/3. The factor 3 (and its reciprocal 1/3) come from tan 60° / tan 30° = √3 / (1/√3) = 3. Claim and reasoning both correct.