Mathematics Standard • Year 11 • Module 2 • Lesson 4
Introduction to Trigonometry
Apply SOHCAHTOA to realistic measurement scenarios — ramps, ladders, surveying, kite strings, and shadow problems.
Problem 1 — Ladder angle (worksite safety)
A 6 m ladder leans against a wall with its base 1.5 m from the wall. WorkSafe recommends ladders be set at an angle between 70° and 80° from the ground.
Set up: What are we solving for?
(i) Label the triangle relative to the angle between the ladder and the ground. Which sides are known: O, A or H? 1 mark
(ii) Use the correct ratio to find the angle the ladder makes with the ground to the nearest degree. 3 marks
(iii) Is the ladder set safely? Justify with a one-sentence comparison to the 70°–80° guideline. 2 marks
Stuck? Revisit lesson § Labelling the Triangle. Ground is adjacent to the angle at the base, ladder is the hypotenuse.Problem 2 — Wheelchair ramp (find the angle)
An accessibility ramp rises 0.6 m vertically over a horizontal run of 9 m. The Australian Standard recommends ramps have an angle of no more than 5° from horizontal.
Set up: What are we solving for?
(i) Identify which two sides are known relative to the angle the ramp makes with the ground. 1 mark
(ii) Calculate the ramp angle to 2 d.p. 3 marks
(iii) Does this ramp comply with the standard? State by how many degrees the ramp is under or over the 5° limit. 2 marks
Stuck? Opposite = 0.6 (vertical rise), Adjacent = 9 (horizontal run) → use tan.Problem 3 — Building shadow (find a length)
A building casts a shadow on flat ground. The sun's elevation angle (angle from ground to the top of the building, measured at the tip of the shadow) is 38°. The shadow is 18 m long.
Set up: What are we solving for?
(i) Sketch the right triangle: building (vertical), shadow (horizontal), line-of-sight to top of building. Which sides are O, A, H relative to the 38° angle? 1 mark
(ii) Find the height of the building to 2 d.p. 3 marks
(iii) Each storey of the building is approximately 3.2 m tall. Estimate the number of storeys (as a whole number). 2 marks
Stuck on (ii)? Building (opposite) is unknown, shadow (adjacent) is known → use tan.Problem 4 — Kite string (find the hypotenuse)
A kite is flying directly above point P on the ground. The string makes an angle of 58° with the ground. The vertical height of the kite above the ground is 24 m (straight up from P).
Set up: What are we solving for?
(i) Identify the right triangle: vertical kite-height, horizontal distance (string-end to P), and the slanted string (hypotenuse). Relative to the 58° angle at the string's end, label the sides. 1 mark
(ii) Calculate the length of the string to 2 d.p. 3 marks
(iii) String is sold in 5 m increments. How long should the kite-flier buy (round UP, justify direction)? 2 marks
Stuck? Vertical height = opposite to 58°; string = hypotenuse → use sin.Problem 5 — Suburban block depth (surveying)
A surveyor stands on the front boundary of a sloping block and looks up at the rear corner. The angle of elevation is 12°, and the slope distance (line-of-sight, hypotenuse) from surveyor to rear corner is 42 m.
Set up: What are we solving for?
(i) Calculate the horizontal (adjacent) distance from the surveyor to a point directly below the rear corner, to 2 d.p. 3 marks
(ii) Calculate the vertical rise (opposite) of the rear corner above the surveyor's level, to 2 d.p. 2 marks
(iii) The block's title document says it is "40 m deep (horizontal)". Compare the title figure to your answer in (i) and write a one-sentence reconciliation. 2 marks
Stuck on (i)? Hypotenuse known, adjacent unknown → use cos.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Ladder angle
Set up. We are finding the angle the ladder makes with the ground, then comparing to a safe range.
(i) Ground (1.5 m) = adjacent; ladder (6 m) = hypotenuse; wall = opposite (unknown).
(ii) A and H → use cos. cos θ = 1.5 / 6 = 0.25. θ = cos⁻¹(0.25) = 75.52...° ≈ 76°.
(iii) 76° is between 70° and 80°, so YES — the ladder is set safely within the recommended range.
Problem 2 — Ramp angle
Set up. We are finding the ramp's slope angle and comparing to the 5° max.
(i) Vertical rise (0.6) = opposite; horizontal run (9) = adjacent.
(ii) O and A → use tan. tan θ = 0.6 / 9 = 0.0667. θ = tan⁻¹(0.0667) = 3.81...° ≈ 3.81° (to 2 d.p.).
(iii) 3.81° < 5°, so the ramp complies with the standard — it is 1.19° under the limit.
Problem 3 — Building shadow
Set up. We are finding the building's height using the shadow length and sun's elevation angle.
(i) Building (vertical, unknown) = opposite; shadow (horizontal, 18 m) = adjacent; line-of-sight = hypotenuse.
(ii) O and A → use tan. tan 38° = h / 18 → h = 18 × tan 38° = 14.062... ≈ 14.06 m (to 2 d.p.).
(iii) Storeys = 14.06 / 3.2 = 4.39 → approximately 4 storeys (round down because there is not enough height for a full 5th storey).
Problem 4 — Kite string
Set up. We are finding the slanted string length given the vertical kite height and string angle.
(i) Vertical height (24 m) = opposite; string = hypotenuse.
(ii) O and H → use sin. sin 58° = 24 / H → H = 24 / sin 58° = 28.298... ≈ 28.30 m (to 2 d.p.).
(iii) Round UP to the next 5 m increment → 30 m of string. Round up because 25 m would be too short (28.30 m needed) and the kite would not reach the height.
Problem 5 — Surveying a sloping block
Set up. We use the slope distance (hypotenuse) and elevation angle to recover horizontal and vertical components.
(i) A and H → use cos. cos 12° = A / 42 → A = 42 × cos 12° = 41.082... ≈ 41.08 m (to 2 d.p.).
(ii) O and H → use sin. sin 12° = O / 42 → O = 42 × sin 12° = 8.733... ≈ 8.73 m (to 2 d.p.).
(iii) The title says 40 m horizontal; my calculation gives 41.08 m. The figures differ by about 1.08 m, likely due to rounding in the angle measurement or in the title's stated depth — the two are broadly consistent. (Acceptable alternative: the title may have been rounded to the nearest metre, in which case 41 m rounds to 40 m? No — 41 rounds to 41. The discrepancy is real and worth flagging.)