Mathematics • Year 10 • Unit 3 • Lesson 6
Combining Pythagoras and Trig in the Real World
Apply Lesson 6 to carpentry braces, square pizzas, kite frames and rooftop ladders. Each problem hides a right-angled triangle — your job is to extract it, decide which tool fits the given information, and solve.
1. Word problems
For each problem: draw the triangle, label what you know, and write down which tool you will use before calculating. A bare answer with no working only earns half marks.
1.1 — Carpenter's brace. A carpenter is building a right-angled triangular brace. The two legs measure 5 cm and 12 cm.
(a) Find the length of the diagonal brace (the hypotenuse).
(b) Find the angle opposite the 5 cm side, to the nearest degree. 3 marks
1.2 — Square pizza box diagonal. A square pizza box has side length 32 cm. A friend wants to know how long the diagonal across the box is.
(a) Find the diagonal in exact (surd) form.
(b) Calculate the diagonal to one decimal place. 3 marks
1.3 — Kite frame angle. A diamond-shaped kite is built around two perpendicular sticks. The horizontal stick is 60 cm and the vertical stick is 80 cm long, so the four outer edges of the kite are each the hypotenuse of a 30-cm-by-40-cm right triangle (each stick is shared between two triangles, halved at the centre).
(a) Find the length of one outer edge of the kite.
(b) Find the angle between the outer edge and the horizontal stick, to the nearest degree. 3 marks
1.4 — Ladder against a wall. A 4 m ladder leans against a vertical wall. The foot of the ladder is 1.5 m away from the wall.
(a) How far up the wall does the ladder reach? Give your answer to two decimal places.
(b) Find the angle the ladder makes with the ground, to the nearest degree. 3 marks
1.5 — TV screen diagonal. A rectangular TV is advertised as having a 165 cm diagonal. The width-to-height ratio is 16:9 (so the width is exactly 16 parts and the height is 9 parts of some unknown length k).
(a) Set up a Pythagoras equation in terms of k: (16k)² + (9k)² = 165².
(b) Solve for k, then find the actual width and height (to the nearest cm). 3 marks
2. Explain your thinking
This question is about reasoning, not just numbers. Use full sentences. 4 marks
2.1 Your classmate says: "If I know the hypotenuse of a right-angled triangle, I should always use trigonometry because the hypotenuse is built into every trig ratio." Using Lesson 6's decision rule, explain (i) when the classmate is correct, (ii) when the classmate is wrong, and (iii) what extra piece of information you would need before reaching for trigonometry. Use the phrase "angle other than 90°" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Carpenter's 5-12 brace
(a) c² = 5² + 12² = 25 + 144 = 169 → c = 13 cm.
(b) sin θ = 5/13 ≈ 0.3846 → θ = sin⁻¹(5/13) ≈ 22.6° → 23° (nearest degree).
This is the classic 5-12-13 Pythagorean triple.
1.2 — 32 cm pizza box diagonal
(a) Diagonal = 32√2 cm (exact).
(b) 32 × √2 ≈ 32 × 1.4142 ≈ 45.3 cm.
Hypotenuse = leg × √2 — the isosceles right triangle shortcut.
1.3 — Kite outer edge and angle
(a) Edge² = 30² + 40² = 900 + 1600 = 2500 → edge = 50 cm. (3-4-5 triple scaled by 10.)
(b) tan θ = 40/30 ≈ 1.333 → θ = tan⁻¹(4/3) ≈ 53° (nearest degree).
1.4 — 4 m ladder, 1.5 m from wall
(a) h² = 4² − 1.5² = 16 − 2.25 = 13.75 → h = √13.75 ≈ 3.71 m.
(b) cos θ = 1.5/4 = 0.375 → θ = cos⁻¹(0.375) ≈ 68° (nearest degree).
Safe ladders typically sit at about 70–75°, so 68° is in the realistic range.
1.5 — 16:9 TV with 165 cm diagonal
(a) (16k)² + (9k)² = 165² → 256k² + 81k² = 27225 → 337k² = 27225.
(b) k² = 27225/337 ≈ 80.79 → k ≈ 8.99.
Width = 16k ≈ 143.8 cm ≈ 144 cm. Height = 9k ≈ 80.9 cm ≈ 81 cm.
Check: 144² + 81² = 20736 + 6561 = 27297 ≈ 165² = 27225 ✓ (small rounding).
2.1 — Explain your thinking (sample response)
(i) The classmate is correct that the hypotenuse appears in two of the three trig ratios (sin = opp/hyp and cos = adj/hyp), so knowing the hypotenuse is useful for trigonometry. (ii) The classmate is wrong to claim trig should "always" be used — if I am only given two sides and need to find the third side, trigonometry has nothing to lock onto and Pythagoras is the only valid tool. (iii) Before reaching for trigonometry I need one extra piece of information: an angle other than 90°. With one angle and one side, SOH CAH TOA works; without one, I must fall back on a² + b² = c².
Marking: 1 for what is correct (hypotenuse appears in sin and cos), 1 for what is wrong (Pythagoras needed when no angle is given), 1 for naming "angle other than 90°", 1 for clear full-sentence structure.