Mathematics • Year 10 • Unit 3 • Lesson 7

Ladders, Shadows, Ramps and Kites

Apply Lesson 7 to the classic right-triangle scenarios: ladder against a wall, shadow cast by a building, wheelchair ramp gradient, kite string at an angle. Use exact values for 30°, 45° and 60° wherever they fit, and always check that the answer makes physical sense.

Apply · Real-World Maths

1. Word problems

For each problem: draw the triangle, label what you know, decide on the strategy, then check the answer is reasonable. Use exact values where 30°, 45° or 60° appears.

1.1 — Ladder lesson-hook. A 5 m ladder leans against a wall at 60° to the ground (the recommended angle for safety).

(a) Using exact values, find how high up the wall the ladder reaches.
(b) Using exact values, find the horizontal distance from the foot of the ladder to the wall.
(c) Convert both answers to decimals (2 d.p.). 3 marks

Stuck? This is exactly the lesson's hook problem. sin 60° = √3/2, cos 60° = 1/2.

1.2 — Tree and shadow. A gum tree casts a shadow 7 m long. The angle from the tip of the shadow up to the top of the tree is 45°.

(a) Without using a calculator, state the height of the tree and justify with an exact value.
(b) Sanity-check your answer: should a tree's shadow at 45° be longer than, equal to, or shorter than the tree? 3 marks

Stuck? At 45°, the opposite and adjacent sides are equal (because tan 45° = 1).

1.3 — Wheelchair ramp. Australian Standard AS 1428.1 says the steepest a wheelchair ramp may be is 1:14 (rise:run). A new ramp must rise 0.5 m to reach a doorway.

(a) Calculate the minimum horizontal run (length along the ground) the ramp must have.
(b) Find the angle the ramp makes with the ground, to the nearest tenth of a degree.
(c) Find the length of the sloping surface of the ramp to two decimal places. 3 marks

Stuck? Run = 14 × rise. For the angle, tan θ = rise/run = 1/14. For the slope, use Pythagoras.

1.4 — Kite string at an angle. A child flies a kite on 50 m of string. The string is taut and makes a 30° angle with the horizontal ground.

(a) Use the exact value sin 30° = 1/2 to find the height of the kite above the ground (assuming the string is held at ground level).
(b) Use the exact value cos 30° = √3/2 to find the horizontal distance from the child to the point directly below the kite. Give an exact value and a decimal to 2 d.p. 3 marks

Stuck? Height is opposite the 30° angle (use sine). Horizontal distance is adjacent (use cosine).

1.5 — Building shadow. The Sydney Tower Eye is approximately 309 m tall. At a certain time of afternoon, it casts a shadow 535 m long across nearby streets.

(a) Find the angle of elevation of the sun to the nearest degree.
(b) Reasonableness check: at noon the sun is nearly overhead so the shadow is short; near sunset it lengthens. Is your angle consistent with a mid-to-late afternoon time? 3 marks

Stuck? tan(angle of elevation) = opposite/adjacent = height/shadow. Use tan⁻¹.

2. Explain your thinking

This question is about reasoning, not just numbers. Use full sentences. 4 marks

2.1 A friend solving an exam question writes "sin 30° = 0.49999". Using Lesson 7's exact-values idea, explain (i) what is correct about this number, (ii) what is mathematically misleading or wrong, (iii) which exact value they should have used instead, and (iv) name one situation where an exact value beats a decimal in a real exam. Use the phrase "exact value" somewhere in your answer.

Stuck? Revisit lesson § "Common Pitfalls" — "Forgetting exact values exist". sin 30° = 1/2 exactly; calculators round and can be in the wrong mode.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — 5 m ladder at 60°

(a) Wall height = 5 × sin 60° = 5 × (√3/2) = 5√3/2 m.
(b) Horizontal distance = 5 × cos 60° = 5 × (1/2) = 5/2 m = 2.5 m.
(c) 5√3/2 ≈ 4.33 m; 5/2 = 2.50 m.
Check via Pythagoras: (5√3/2)² + (5/2)² = 75/4 + 25/4 = 100/4 = 25 = 5² ✓.

1.2 — 7 m shadow, 45° angle of elevation

(a) tan 45° = height/shadow = 1, so height = shadow = 7 m.
(b) At 45°, shadow length equals tree height (because tan 45° = 1 means opposite = adjacent).
This is the "shadow at 45° = your height" trick used by surveyors.

1.3 — Wheelchair ramp, rise 0.5 m at 1:14

(a) Run = 14 × 0.5 = 7 m.
(b) tan θ = 0.5/7 = 1/14 ≈ 0.0714 → θ = tan⁻¹(1/14) ≈ 4.1°.
(c) Slope = √(0.5² + 7²) = √(0.25 + 49) = √49.25 ≈ 7.02 m.
A 1:14 ramp is very gentle — about 4°. Steeper ramps fail the access standard.

1.4 — Kite at 30°, 50 m of string

(a) Height = 50 × sin 30° = 50 × (1/2) = 25 m.
(b) Horizontal distance = 50 × cos 30° = 50 × (√3/2) = 25√3 m ≈ 43.30 m.
Check: 25² + (25√3)² = 625 + 1875 = 2500 = 50² ✓.

1.5 — Sydney Tower Eye, 309 m tall, 535 m shadow

(a) tan(elevation) = 309/535 ≈ 0.5776 → elevation = tan⁻¹(0.5776) ≈ 30°.
(b) Mid-to-late afternoon, the sun's elevation is well below noon's near-vertical position. 30° is consistent with a few hours before sunset. ✓ Reasonable.

2.1 — Explain your thinking (sample response)

(i) The number 0.49999 is correct as a rounded decimal — a calculator working to four significant figures will produce something very close to 0.5 for sin 30°. (ii) It is misleading because it suggests the value is irrational or messy, when in fact sin 30° equals exactly 0.5 — there is no rounding involved. Writing 0.49999 in an exam invites compounding rounding errors in the next step. (iii) The friend should have used the exact value sin 30° = 1/2. (iv) Exact values matter in any problem where a later step multiplies or squares the trig value — for example, computing the area of a triangle using (1/2) × base × height × sin(angle), where a rounding error of 0.00001 can become a few centimetres in the final answer.

Marking: 1 for noting the decimal is close to correct, 1 for explaining why sin 30° = 1/2 exactly, 1 for using the phrase "exact value", 1 for a clear example of where exact values matter in an exam.