Mathematics • Year 9 • Unit 4 • Lesson 3
Solving Triangles in the Real World
Use trig to solve real triangles — a roof pitch, a wheelchair ramp, a flying kite, a treadmill incline, and a fence diagonal. Pick the right ratio, set up, rearrange, solve, and check. Then explain your method in your own words.
1. Word problems
Each problem asks you to solve a right-angled triangle. Draw a sketch, label opposite / adjacent / hypotenuse, pick the ratio, set up, and solve. Calculator in degree mode. Round final answers to 2 dp unless told otherwise.
1.1 — Roof pitch. A house roof rises 1.8 m above the ceiling line over a horizontal half-span of 4.5 m (from the ridge straight down to the wall plate). Find the roof pitch (the angle between the roof and horizontal) to the nearest degree. 3 marks
1.2 — Wheelchair ramp. The Australian Standard for wheelchair ramps requires the angle of inclination to be no more than 4.8° (about 1:12 gradient). A new ramp at a community hall has a horizontal length of 6 m and rises 0.5 m vertically.
(a) Find the actual angle of inclination.
(b) Does this ramp meet the standard? 3 marks
1.3 — Kite in the sky. A child flies a kite on a 50 m string. The string makes an angle of 40° with the horizontal ground.
(a) How high is the kite above the ground? (Ignore the height of the child's hand.)
(b) How far is the kite, measured horizontally, from the child? 3 marks
1.4 — Treadmill incline. A treadmill's belt is 1.4 m long (the slope, i.e. the hypotenuse). When set at an incline of 7°, the back end of the belt sits on the ground and the front end is lifted.
(a) How high is the front of the belt lifted above the ground? (Convert to cm.)
(b) Verify your answer using cos and Pythagoras (find the horizontal distance, then check opp² + adj² = hyp²). 3 marks
1.5 — Fence diagonal. A rectangular paddock is 30 m long and 15 m wide. A farmer wants to know the length of the diagonal fence and the angle the diagonal makes with the longer side.
(a) Find the diagonal length to 1 dp (Pythagoras).
(b) Find the angle the diagonal makes with the 30 m side, to 1 dp. 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate is solving the kite problem (1.3) and writes: "sin 40° = 50 / x, so x = 50 / sin 40° ≈ 77.8 m, so the kite is 77.8 m up." In your own words, explain (i) what they got mixed up about which side is the hypotenuse, (ii) why their answer can't be right (think about the kite's height vs the string length), and (iii) what the correct equation and answer should be. Refer to "the hypotenuse is the longest side" somewhere.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Roof pitch
tan θ = rise / half-span = 1.8 / 4.5 = 0.4 → θ = tan⁻¹(0.4) ≈ 22°.
A typical Australian house roof pitch is between 20° and 30°, so this is realistic.
1.2 — Wheelchair ramp
(a) tan θ = 0.5 / 6 = 0.0833 → θ = tan⁻¹(0.0833) ≈ 4.76°.
(b) The standard allows up to 4.8°, and 4.76° is just under 4.8° → the ramp meets the standard (only just).
This is a tight pass — anything steeper would need to be redesigned.
1.3 — Kite in the sky
Hypotenuse = string = 50 m. Angle with ground = 40°.
(a) Height = 50 × sin 40° = 50 × 0.6428 ≈ 32.14 m.
(b) Horizontal distance = 50 × cos 40° = 50 × 0.7660 ≈ 38.30 m.
Sanity check: both are less than the 50 m string, as they must be.
1.4 — Treadmill incline
(a) Rise = 1.4 × sin 7° = 1.4 × 0.1219 ≈ 0.1706 m = 17.06 cm (or about 17 cm).
(b) Horizontal = 1.4 × cos 7° ≈ 1.4 × 0.9925 ≈ 1.3895 m. Check: 0.1706² + 1.3895² ≈ 0.0291 + 1.9307 ≈ 1.9598 ≈ 1.96 = 1.4². ✓
1.5 — Fence diagonal
(a) Diagonal = √(30² + 15²) = √(900 + 225) = √1125 ≈ 33.5 m.
(b) cos θ = 30 / 33.5 ≈ 0.8944 → θ = cos⁻¹(0.8944) ≈ 26.6°.
Alternative: tan θ = 15/30 = 0.5 → θ = tan⁻¹(0.5) ≈ 26.6°. ✓ Both methods agree.
2.1 — Explain your thinking (sample response)
My classmate mixed up which side is the hypotenuse. In the kite triangle, the string (50 m) IS the hypotenuse — it's the long sloping line from the child's hand to the kite. The height of the kite is OPPOSITE the 40° angle, not the hypotenuse. So the correct equation is sin 40° = height / 50, not sin 40° = 50 / height. Their answer of 77.8 m can't be right because the hypotenuse is the longest side of a right-angled triangle, so the height (opposite) can never be greater than 50 m. The correct working is height = 50 × sin 40° ≈ 32.14 m, which is less than 50 — exactly as it should be.
Marking: 1 mark for identifying the string as the hypotenuse; 1 for explaining height ≤ string; 1 for the correct equation; 1 for a clear, full-sentence explanation with the right final answer.