Mathematics • Year 9 • Unit 3 • Lesson 6
SOH–CAH–TOA — Mixed Challenge
Pull together everything from L6: identifying sides relative to θ, picking the right ratio, simplifying fractions, and reasoning about similar triangles. Six mixed problems, one mistake-spotter, one open-ended puzzle.
1. Mixed problems — choose the right ratio
Decide which of sin, cos or tan applies before you start writing. Show your working. 3 marks each
1.1 A right triangle has opp = 8, adj = 15. Find tan θ as a fraction in simplest form, and find sin θ as a fraction (use Pythagoras to find hyp first).
1.2 A right triangle has hyp = 25 and opp = 7. Find cos θ as a fraction (Pythagoras gives adj = 24).
1.3 In a 5–12–13 right triangle, θ is the angle opposite the side of length 5. Find sin θ, cos θ and tan θ as fractions. Then state which one is bigger: cos θ or sin θ, and why this makes sense for an angle smaller than 45°.
1.4 Two right triangles share the acute angle θ. Triangle P has opp = 12 and hyp = 20. Triangle Q has hyp = 35. Find the opp side in Triangle Q without using a calculator (use the fact that sin θ is the same in both).
1.5 A right triangle has cos θ = 4/5. Without finding the angle, find sin θ and tan θ. (Hint: this matches a 3–4–5 family triangle. Use Pythagoras with any convenient pair.)
1.6 A right triangle has opp = 5 and adj = 5 (so it's a 45°-45°-90° isosceles right triangle). Find tan θ exactly, then find sin θ and cos θ in terms of √2 (use Pythagoras: hyp = 5√2).
2. Find the mistake
A Year 9 student has tried to find sin θ, cos θ and tan θ for a right triangle with opp = 6, adj = 8, hyp = 10. Exactly one line contains a Year 9 trig mistake (not an arithmetic slip — a confusion of the rule itself). Spot it, explain why it's wrong, and re-do the working correctly. 3 marks
Student's working — find the three ratios:
Line 1: opp = 6, adj = 8, hyp = 10.
Line 2: sin θ = opp/hyp = 6/10 = 0.6
Line 3: cos θ = adj/hyp = 8/10 = 0.8
Line 4: tan θ = opp/hyp = 6/10 = 0.6
(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, including the corrected final answer.
Stuck? Recall TOA — Tangent uses Opposite over Adjacent, never the hypotenuse.3. Open-ended challenge — design your own triangle
This question has more than one valid answer. 4 marks
3.1 Design two different right triangles whose sides are all whole numbers AND whose acute angle θ satisfies tan θ = 3/4. Both triangles must have an angle θ such that tan θ is exactly 3/4 but the two triangles must NOT be the same size.
For each triangle:
(i) Write the three side lengths (opp, adj, hyp).
(ii) Verify tan θ = 3/4 from your numbers.
(iii) Verify the sides satisfy Pythagoras (opp² + adj² = hyp²).
Bonus reasoning: Explain in one sentence why both triangles MUST have the same sin θ and cos θ (not just the same tan θ).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — opp = 8, adj = 15
tan θ = 8/15 (already in simplest form; gcd(8,15)=1).
Pythagoras: hyp² = 8² + 15² = 64 + 225 = 289, so hyp = 17.
sin θ = opp/hyp = 8/17. This is an 8–15–17 Pythagorean triple.
1.2 — hyp = 25, opp = 7
Pythagoras: adj² = 25² − 7² = 625 − 49 = 576, so adj = 24.
cos θ = adj/hyp = 24/25. 7–24–25 triple.
1.3 — 5–12–13, θ opposite the 5
opp = 5, adj = 12, hyp = 13. sin θ = 5/13; cos θ = 12/13; tan θ = 5/12.
cos θ (12/13 ≈ 0.92) is BIGGER than sin θ (5/13 ≈ 0.38). This makes sense because for angles smaller than 45°, the opp is shorter than the adj, so opp/hyp < adj/hyp. (At exactly 45°, sin = cos.)
1.4 — Same θ, different size
In Triangle P: sin θ = 12/20 = 3/5.
In Triangle Q: sin θ must also be 3/5 (same angle). So opp/35 = 3/5, giving opp = 35 × 3/5 = 21.
Triangle Q is just Triangle P scaled by 35/20 = 1.75 — all sides multiplied by 1.75.
1.5 — Given cos θ = 4/5
cos θ = adj/hyp = 4/5 matches a 3–4–5 family triangle with adj = 4, hyp = 5, opp = 3 (by Pythagoras: 5² − 4² = 9, so opp = 3).
sin θ = opp/hyp = 3/5. tan θ = opp/adj = 3/4.
The actual size doesn't matter — any 3–4–5 scaling gives the same ratios.
1.6 — 45°-45°-90° (opp = adj = 5)
tan θ = opp/adj = 5/5 = 1 exactly.
hyp = √(5² + 5²) = √50 = 5√2.
sin θ = 5/(5√2) = 1/√2 = √2/2 ≈ 0.707.
cos θ = 5/(5√2) = 1/√2 = √2/2 ≈ 0.707.
For 45°, sin and cos are equal — this is the only acute angle where that happens.
2 — Find the mistake
(a) The mistake is on Line 4.
(b) The student has used opp/hyp for tan, which is actually the formula for sin (SOH). The correct formula for tan is opp/adj (TOA) — tangent NEVER uses the hypotenuse.
(c) Corrected: tan θ = opp/adj = 6/8 = 3/4 = 0.75.
This is the exact "Using hyp in tan" pitfall flagged in the lesson's Common Pitfalls card. Memory aid: TOA — opposite on TOP, adjacent on BOTTOM, no hyp anywhere.
3 — Open-ended challenge (sample solution)
We need tan θ = 3/4, so opp/adj = 3/4. The simplest whole-number triangle is the 3–4–5. Pythagoras gives hyp = 5. Any scaling of 3–4–5 by a whole number keeps tan θ = 3/4.
Triangle 1: opp = 3, adj = 4, hyp = 5.
(ii) tan θ = 3/4 ✓.
(iii) 3² + 4² = 9 + 16 = 25 = 5² ✓.
Triangle 2: opp = 9, adj = 12, hyp = 15 (a 3× scaling of triangle 1).
(ii) tan θ = 9/12 = 3/4 ✓.
(iii) 9² + 12² = 81 + 144 = 225 = 15² ✓.
Other valid answers: (6, 8, 10), (12, 16, 20), (15, 20, 25), … — any k×(3, 4, 5) with whole k ≥ 1.
Bonus reasoning: Same angle θ implies the triangles are similar, and similar triangles have ALL trig ratios identical — not just tan. Both triangles have sin θ = 3/5 = 0.6 and cos θ = 4/5 = 0.8.
Marking: 1 mark per triangle (must verify tan and Pythagoras); 1 mark for naming the second as a non-trivial scaling of the first; 1 mark for the bonus reasoning.