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.)