Mathematics Standard • Year 11 • Module 2 • Lesson 14
Right-Angled Trig: Finding Sides — Problem Set
Apply SOHCAHTOA to ladders, ramps, surveying, navigation and construction scenarios — draw the triangle, label, pick the ratio, solve.
Problem 1 — Painter's ladder safety
A 6 m extension ladder leans against a vertical wall. The ladder makes an angle of 72° with the (horizontal) ground.
Set up: What are we solving for?
(i) Sketch and label the triangle (right angle at the base of the wall, 72° at ground, ladder = H, wall reach = O, base distance = A). 1 mark
(ii) Find how high up the wall the ladder reaches, to 2 d.p. 2 marks
(iii) Find the distance from the base of the wall to the foot of the ladder, to 2 d.p. 1 mark
Stuck? Revisit lesson § Worked Example 1. Wall reach uses sin (O and H); base distance uses cos (A and H).Problem 2 — Surveyor and a cliff (degrees and minutes)
A surveyor stands 80 m from the base of a vertical cliff on level ground. She measures the angle of elevation to the top of the cliff as 38°42'.
Set up: What are we solving for?
(i) Convert 38°42' to decimal degrees. 1 mark
(ii) Find the height of the cliff, to the nearest metre. 2 marks
(iii) If she now stands 40 m from the base (half the distance), and the cliff is the same height as in (ii), would the angle of elevation be double, more than double, or less than double 38°42'? Justify in one sentence (you don't need to calculate the new angle exactly). 2 marks
Stuck on (iii)? tan is non-linear: doubling the height (or halving the base) does NOT double the angle.Problem 3 — Disability-access ramp design
The Australian Standard AS 1428.1 recommends a maximum slope of 1 in 14 for an accessible ramp — equivalent to an angle of about 4.1° with the horizontal. A school needs a ramp to rise 0.75 m to a door, and must conform to a slope angle of 4°.
Set up: What are we solving for?
(i) Sketch a labelled triangle (rise = O = 0.75 m, angle = 4° at the ground, run = A, ramp surface = H). 1 mark
(ii) Find the horizontal run A (the floor distance covered by the ramp), correct to 2 d.p. 2 marks
(iii) Find the length of the ramp surface H, correct to 2 d.p. 2 marks
Stuck? For (ii) use tan (O and A); for (iii) use sin (O and H).Problem 4 — Yacht navigation (bearing → distance)
A yacht leaves port and sails on a bearing such that its straight-line path makes an angle of 25° east of north for 4.8 km, then drops anchor.
Set up: What are we solving for?
(i) Sketch a right-angled triangle with the right angle at the port "ground": the 4.8 km path is the hypotenuse, the north side (A) and the east side (O) are the legs, and the 25° angle is at the port. 1 mark
(ii) Find how far north of port the anchorage is, to 2 d.p. (km). 2 marks
(iii) Find how far east of port the anchorage is, to 2 d.p. (km). 2 marks
Stuck? With the 25° angle measured from north, the north side is adjacent (cos) and the east side is opposite (sin).Problem 5 — Roof rafter cost
A simple gable roof has a horizontal half-span of 4.5 m. The roof pitch (angle of the rafter to the horizontal) is 27°. Treated pine rafters cost $32 per metre.
Set up: What are we solving for?
(i) Sketch the right-angled triangle (half-span = A = 4.5 m, rise = O, rafter = H, angle 27° at the eave). 1 mark
(ii) Find the rise of the roof (top of the ridge above the eave) to 2 d.p. 1 mark
(iii) Find the length of one rafter (the H side) to 2 d.p. 2 marks
(iv) The roof has 14 such rafters (7 on each side). Find the total rafter cost to the nearest dollar. 2 marks
Stuck? Rafter length needs H from cos 27° = A ÷ H, i.e. H = 4.5 ÷ cos 27°.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Ladder at 72°
Set up. Ladder is H = 6 m, angle 72° at ground; wall reach is O; base distance is A.
(i) Diagram: right angle at base of wall; 72° at ground end; 6 m is the hypotenuse; vertical wall reach O and horizontal base A labelled.
(ii) O = 6 × sin 72° ≈ 6 × 0.9511 ≈ 5.71 m.
(iii) A = 6 × cos 72° ≈ 6 × 0.3090 ≈ 1.85 m.
Problem 2 — Surveyor 80 m from cliff
Set up. Right angle at the base of the cliff; angle 38°42' at the surveyor; A = 80 m; O = cliff height.
(i) 38°42' = 38 + 42/60 = 38.7°.
(ii) h = 80 × tan 38.7° ≈ 80 × 0.8002 ≈ 64.02 ≈ 64 m.
(iii) Halving the base (A) at the same height (O) gives tan θ' = 64/40 = 1.6 → θ' ≈ 57.99°, which is less than double 38.7° (≈77.4° would be double). Because tan is a non-linear function, halving the base distance does not produce a doubling of the angle.
Problem 3 — Accessible ramp, rise 0.75 m, slope 4°
Set up. O = 0.75 m, angle 4°; need A (run) and H (ramp surface).
(i) Diagram: right angle at the foot of the rise; 4° at the ground end; rise 0.75 m vertical; run A horizontal; ramp H sloped.
(ii) tan 4° = 0.75 ÷ A ⇒ A = 0.75 ÷ tan 4° ≈ 0.75 ÷ 0.0699 ≈ 10.72 m.
(iii) sin 4° = 0.75 ÷ H ⇒ H = 0.75 ÷ sin 4° ≈ 0.75 ÷ 0.0698 ≈ 10.75 m.
Problem 4 — Yacht N25°E for 4.8 km
Set up. Right angle at port "axes" junction; H = 4.8 km path; angle 25° from north; A = north distance (adjacent), O = east distance (opposite).
(i) Diagram: north-pointing leg A, east-pointing leg O, hypotenuse 4.8 km, 25° at port between N and the path.
(ii) North distance: A = 4.8 × cos 25° ≈ 4.8 × 0.9063 ≈ 4.35 km.
(iii) East distance: O = 4.8 × sin 25° ≈ 4.8 × 0.4226 ≈ 2.03 km.
Problem 5 — Gable roof rafters
Set up. A = 4.5 m, angle 27° at the eave; need O (rise) and H (rafter length); 14 rafters total.
(i) Diagram: right angle at the ridge directly above one eave; A = 4.5 m horizontal; O = rise vertical; H = sloped rafter; 27° at the eave.
(ii) Rise = 4.5 × tan 27° ≈ 4.5 × 0.5095 ≈ 2.29 m.
(iii) Rafter H = 4.5 ÷ cos 27° ≈ 4.5 ÷ 0.8910 ≈ 5.05 m.
(iv) Total rafter length = 14 × 5.05 = 70.70 m. Cost = 70.70 × $32 = $2,262.40 ≈ $2,262.