Mathematics • Year 10 • Unit 3 • Lesson 6
Pythagoras and Trigonometry Combined — Mixed Challenge
Pull together every idea from Lesson 6: decide between Pythagoras, trigonometry and both-in-sequence; use exact values for 45°; spot a classmate's plausible mistake; then design your own two-step right-triangle problem.
1. Mixed problems — choose the right tool
Each question uses a different idea from Lesson 6. Decide which tool applies before you start writing. Show your working. 3 marks each
1.1 A right-angled triangle has legs 7 cm and 24 cm. Find the hypotenuse and identify the Pythagorean triple.
1.2 A right-angled triangle has hypotenuse 14 cm and one acute angle of 35°. Find both legs (to 2 d.p.).
1.3 An isosceles right triangle has equal sides of 8 cm. Use exact values to write the hypotenuse and to verify that sin 45° = 1/√2.
1.4 A right-angled triangle has legs 8 cm and 15 cm. Find both acute angles (to 1 d.p.) and verify they sum to 90°.
1.5 A 6 m ladder leans against a wall at exactly 45° to the ground. Use exact values (no calculator) to find (a) how far up the wall it reaches, (b) the horizontal distance from the foot to the wall.
1.6 A right-angled triangle has hypotenuse 17 cm and one leg 8 cm. Find the third side (using Pythagoras) and the angle opposite the 8 cm side (to 1 d.p.).
2. Find the mistake
Another Year 10 student wants to find the angle opposite the 6 cm side in a right triangle with legs 6 cm and 8 cm. Their working is below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — find θ opposite the 6 cm leg:
Line 1: Hypotenuse c² = 6² + 8² = 36 + 64 = 100, so c = 10 cm.
Line 2: θ is opposite the 6 cm leg, hypotenuse is 10, so use cosine: cos θ = 6/10 = 0.6.
Line 3: θ = cos⁻¹(0.6) ≈ 53.13°.
Line 4: So the angle opposite the 6 cm side is ≈ 53.1°.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? SOH CAH TOA — which letter pairs opposite with hypotenuse? Cosine pairs adjacent with hypotenuse, not opposite.3. Open-ended challenge — design your own two-step problem
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design a real-world right-angled triangle problem (e.g. flagpole shadow, sail on a boat, picture frame diagonal) that requires both Pythagoras and trigonometry in sequence to solve. Your problem must:
(i) Start from two side lengths only (no angle given).
(ii) Ask for an angle in the final answer (so inverse trig is needed).
(iii) Use a Pythagorean triple (3-4-5, 5-12-13, 8-15-17, or 7-24-25) so the intermediate hypotenuse is a clean whole number.
Then solve your own problem, showing both steps clearly. End with a sanity check: does the angle sit between 0° and 90°?
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Legs 7 and 24
c² = 49 + 576 = 625 → c = 25 cm. Pythagorean triple: 7-24-25.
1.2 — Hypotenuse 14, angle 35°
Opposite leg = 14 × sin 35° ≈ 14 × 0.5736 ≈ 8.03 cm.
Adjacent leg = 14 × cos 35° ≈ 14 × 0.8192 ≈ 11.47 cm.
Check: 8.03² + 11.47² ≈ 64.5 + 131.6 = 196 = 14² ✓.
1.3 — Isosceles right with legs 8
c² = 8² + 8² = 128 → c = √128 = 8√2 cm.
Verify: sin 45° = opposite/hypotenuse = 8/(8√2) = 1/√2 ✓.
1.4 — Legs 8 and 15
Hypotenuse: c² = 64 + 225 = 289 → c = 17 cm (8-15-17 triple).
Angle opposite 8: sin θ = 8/17 → θ ≈ 28.1°.
Angle opposite 15: sin φ = 15/17 → φ ≈ 61.9°.
Sum: 28.1 + 61.9 = 90° ✓.
1.5 — 6 m ladder at 45° (exact values)
(a) Height = 6 × sin 45° = 6 × (√2/2) = 3√2 m (≈ 4.24 m).
(b) Horizontal distance = 6 × cos 45° = 6 × (√2/2) = 3√2 m (≈ 4.24 m).
At 45° both legs are equal — the ladder reaches as high as it sits out.
1.6 — Hypotenuse 17, leg 8
Third side: a² = 17² − 8² = 289 − 64 = 225 → a = 15 cm (8-15-17 again).
Angle opposite 8: sin θ = 8/17 → θ = sin⁻¹(8/17) ≈ 28.1°.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The angle θ is opposite the 6 cm side, not adjacent. Cosine pairs adjacent with hypotenuse (CAH), but here we have opposite and hypotenuse, so the correct ratio is sine (SOH).
(c) Corrected working:
Hypotenuse: c² = 6² + 8² = 100, so c = 10 cm.
sin θ = opposite/hypotenuse = 6/10 = 0.6.
θ = sin⁻¹(0.6) ≈ 36.9°.
Trap: the student got the right number (0.6) but paired it with the wrong ratio, so the angle came out as the complement (53.1° instead of 36.9°).
3 — Open-ended challenge (sample solution)
Problem. A sailmaker cuts a right-angled triangular sail with legs of 5 m and 12 m. They want to fold the long edge over for hemming — but first they need to know (a) how long that long edge is and (b) the angle between the long edge and the 12 m leg.
Step 1 (Pythagoras): c² = 5² + 12² = 25 + 144 = 169 → c = 13 m (uses the 5-12-13 triple).
Step 2 (inverse trig): The angle between the 13 m long edge and the 12 m leg sits opposite the 5 m leg. tan θ = 5/12 → θ = tan⁻¹(5/12) ≈ 22.6°.
Sanity check: 22.6° is between 0° and 90° ✓. The other acute angle is ≈ 67.4°, and 22.6 + 67.4 + 90 = 180 ✓.
Marking: 1 for a plausible real-world setup, 1 for using a Pythagorean triple, 1 for the Pythagoras step, 1 for the correct inverse-trig step with sanity check. Full marks for any valid two-step problem that meets all three rules.