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.
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.