Mathematics • Year 10 • Unit 3 • Lesson 1
Trig Ratios — Mixed Challenge
Pull together every idea from Lesson 1: labelling H, O, A relative to a marked angle θ, writing the three trig ratios as fractions, recalling the exact values for 30°, 45° and 60°, and spotting the classic mislabelling mistake. Then build your own scaled Pythagorean triple.
1. Mixed problems — choose the right idea
Each question pulls on a different idea from Lesson 1. Decide whether you need to label sides, write a ratio, or use an exact value before you start writing. 2-3 marks each
1.1 A right-angled triangle has hypotenuse 41, with the side opposite θ equal to 9. The triangle is part of the 9-40-41 Pythagorean triple. Write sin θ, cos θ and tan θ as fractions in simplest form. 3 marks
1.2 Without a calculator, evaluate sin² 30° + cos² 30°. (Reminder: sin² 30° means (sin 30°)².) Give your answer as a single number. 2 marks
1.3 In a right-angled triangle, cos θ = 4/5. Sketch a possible triangle (label H, O and A) and use Pythagoras to find the opposite side as a fraction of H, then write tan θ. 3 marks
1.4 Without a calculator, evaluate: 2 × sin 45° × cos 45°. Give your answer as a single number. 2 marks
1.5 The same right-angled triangle has two marked angles θ (at one corner) and α (at the other non-right corner). Given sin θ = 5/13, write sin α and cos α as fractions. (Hint: from α's point of view, what was the opposite for θ becomes the adjacent.) 3 marks
1.6 A triangle has sides 8, 15, 17. Verify it is right-angled using Pythagoras, then write the three trig ratios for the angle opposite the side of length 8. 3 marks
2. Find the mistake
Another Year 10 student has tried to write the trig ratios for the triangle below. Their working is shown. Exactly one line contains a mistake. Spot it, explain why it is wrong, then re-do the working correctly. 3 marks
Triangle: right angle at top-left, angle θ at bottom-left. H = 13, side opposite θ = 5, side adjacent to θ = 12. Student's working:
Line 1: H = 13, O = 5, A = 12.
Line 2: sin θ = O / H = 5 / 13.
Line 3: cos θ = H / A = 13 / 12.
Line 4: tan θ = O / A = 5 / 12.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected line in full.
Stuck? Each of the three ratios must be a fraction less than or equal to 1 when the hypotenuse is on the bottom. Check whether 13/12 is even possible for cos θ in a right-angled triangle.3. Open-ended challenge — design your own triangle
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design two different right-angled triangles (with integer side lengths only) such that the angle θ has tan θ = 3/4 in both triangles. The two triangles must not be identical — they must be different sizes.
For each triangle you design:
(i) State H, O and A as integer side lengths.
(ii) Verify with Pythagoras that the triangle is right-angled.
(iii) Confirm that tan θ = O / A simplifies to 3/4.
Bonus: Explain in one sentence why any triangle with tan θ = 3/4 must have the same sin θ and cos θ values.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 9-40-41 triangle
H = 41, O = 9, A = 40. sin θ = 9/41, cos θ = 40/41, tan θ = 9/40.
None of these simplify — gcd(9, 41) = gcd(40, 41) = gcd(9, 40) = 1.
1.2 — sin² 30° + cos² 30°
sin 30° = 1/2, so sin² 30° = 1/4.
cos 30° = √3 / 2, so cos² 30° = 3/4.
Sum = 1/4 + 3/4 = 1.
This is no coincidence — sin² θ + cos² θ = 1 for every angle. You will meet it formally in senior trig.
1.3 — cos θ = 4/5
cos θ = A / H, so we can take A = 4 and H = 5.
By Pythagoras, O² = H² − A² = 25 − 16 = 9, so O = 3.
Therefore tan θ = O / A = 3/4. (Triangle: 3-4-5.)
1.4 — 2 × sin 45° × cos 45°
sin 45° = cos 45° = 1/√2.
2 × (1/√2) × (1/√2) = 2 / 2 = 1.
Another preview of senior trig: 2 sin θ cos θ = sin (2θ), and sin 90° = 1. ✓
1.5 — Same triangle, complementary angle α
Given sin θ = 5/13, the triangle is 5-12-13 (Pythagorean triple).
From α's point of view, the side that was opposite θ is now adjacent, and vice versa. The hypotenuse does not change.
So sin α = (12/13) and cos α = (5/13).
sin α = 12/13, cos α = 5/13. Note sin α = cos θ and cos α = sin θ — the complementary angle pattern.
1.6 — 8-15-17 triangle
Pythagoras check: 8² + 15² = 64 + 225 = 289 = 17². ✓ Right-angled.
For the angle opposite 8: H = 17, O = 8, A = 15.
sin θ = 8/17, cos θ = 15/17, tan θ = 8/15.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student has inverted the cosine ratio. SOH-CAH-TOA says C-A-H: Cosine = Adjacent ÷ Hypotenuse, so the hypotenuse goes on the bottom, not the top. Also, cos θ can never exceed 1 in a right-angled triangle, but 13/12 ≈ 1.083, which is impossible.
(c) Corrected: cos θ = A / H = 12 / 13.
The "Spot the Trap" card in Lesson 1 calls this inversion out explicitly.
3 — Open-ended (sample solutions)
tan θ = 3/4 means O : A = 3 : 4 (and so the triangle is similar to a 3-4-5).
Triangle A — scale ×1: O = 3, A = 4, H = 5. Pythagoras: 3² + 4² = 9 + 16 = 25 = 5². ✓ tan θ = 3/4. ✓
Triangle B — scale ×3: O = 9, A = 12, H = 15. Pythagoras: 9² + 12² = 81 + 144 = 225 = 15². ✓ tan θ = 9/12 = 3/4. ✓
Bonus: Any two right-angled triangles that share the same θ are similar — every side scales by the same factor. So the ratios sin θ, cos θ, tan θ depend only on θ, not on the size. (sin θ = 3/5 and cos θ = 4/5 in both triangles above.)
Marking: 1 mark for each correct triangle (× 2 = 2 marks), 1 mark for a correct Pythagoras check on each, and 1 mark for the similarity explanation in the bonus. Any pair of scaled 3-4-5 triangles is acceptable.