Mathematics • Year 9 • Unit 3 • Lesson 6
Introducing Trigonometric Ratios
Lock in SOH–CAH–TOA: sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj. Build from one fully-worked example to a guided fill-in to eight independent ratio problems.
1. I do, fully worked example
Read every line. Each step shows how to label the triangle before picking the right ratio.
Problem. A right triangle has opp = 3, adj = 4, hyp = 5 with respect to angle θ. Find sin θ, cos θ and tan θ as exact fractions and as decimals.
Step 1, Mark θ and label the three sides.
opp = 3 (across from θ), adj = 4 (next to θ, not the hypotenuse), hyp = 5 (across from the right angle, always the longest).
Reason: the right angle has the box symbol; the hyp is opposite it. The opp is across from θ; the adj is the remaining leg.
Step 2, Sine uses opp and hyp (SOH).
sin θ = opp/hyp = 3/5 = 0.6
Reason: SOH = Sine, Opposite over Hypotenuse.
Step 3, Cosine uses adj and hyp (CAH).
cos θ = adj/hyp = 4/5 = 0.8
Reason: CAH = Cosine, Adjacent over Hypotenuse.
Step 4, Tangent uses opp and adj (TOA).
tan θ = opp/adj = 3/4 = 0.75
Reason: TOA = Tangent, Opposite over Adjacent. Notice no hyp.
Answer: sin θ = 3/5 = 0.6, cos θ = 4/5 = 0.8, tan θ = 3/4 = 0.75.
2. We do, fill in the missing steps
A right triangle has opp = 5, adj = 12, hyp = 13 with respect to angle θ. Fill in each blank line. 4 marks
Step 1, Identify sides. opp = ____, adj = ____, hyp = ____ .
Step 2, Apply SOH:
sin θ = opp/hyp = ____ / ____ = ____
Step 3, Apply CAH:
cos θ = adj/hyp = ____ / ____ = ____
Step 4, Apply TOA:
tan θ = opp/adj = ____ / ____ = ____
Step 5, Sense check. sin θ and cos θ should both be between ____ and ____ for an acute angle. Is tan θ allowed to be bigger than 1? ____
3. You do, independent practice
Show your working under each problem. Foundation (3.1–3.4) gives you the side lengths. Standard (3.5–3.6) asks you to choose the right ratio. Extension (3.7–3.8) tests the size-doesn't-matter idea.
Foundation, read off one ratio
3.1 A right triangle has opp = 6, hyp = 10. Find sin θ as a fraction in simplest form and as a decimal. 1 mark
3.2 A right triangle has adj = 8, hyp = 17. Find cos θ as a fraction. 1 mark
3.3 A right triangle has opp = 7, adj = 24. Find tan θ as a fraction. 1 mark
3.4 Without a calculator, state which trig ratio CANNOT use the hypotenuse, and write its formula. 1 mark
Standard, choose the right ratio
3.5 A right triangle has opp = 8, adj = 6, hyp = 10. Find all three of sin θ, cos θ and tan θ as decimals. 2 marks
3.6 A right triangle has opp = 9, adj = 40. State which ratio links these two sides only (no hyp), then compute it. 2 marks
Extension, push your thinking
3.7 Triangle A has opp = 6, hyp = 10. Triangle B has opp = 15, hyp = 25. Show that sin θ is exactly the same in both, and explain in one sentence why. 3 marks
3.8 A right triangle has opp = 9, hyp = 15. Find cos θ (as a fraction) WITHOUT using a calculator. Hint: use Pythagoras to find adj first. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (5–12–13 triangle)
Step 1: opp = 5, adj = 12, hyp = 13.
Step 2: sin θ = 5/13 ≈ 0.385.
Step 3: cos θ = 12/13 ≈ 0.923.
Step 4: tan θ = 5/12 ≈ 0.417.
Step 5: sin and cos sit between 0 and 1 for acute angles. Tan CAN be bigger than 1, yes.
3.1, sin θ with opp = 6, hyp = 10
sin θ = 6/10 = 3/5 = 0.6.
3.2, cos θ with adj = 8, hyp = 17
cos θ = adj/hyp = 8/17 ≈ 0.471.
3.3, tan θ with opp = 7, adj = 24
tan θ = opp/adj = 7/24 ≈ 0.292.
3.4, Ratio without hyp
Tangent: tan θ = opp/adj. It uses only the two legs, never the hypotenuse.
3.5, All three ratios (6–8–10)
sin θ = 8/10 = 0.8. cos θ = 6/10 = 0.6. tan θ = 8/6 ≈ 1.33.
Note: this is a 3–4–5 triangle scaled ×2 with θ at the larger acute angle. Tan exceeds 1 because opp > adj.
3.6, opp + adj, no hyp
The two legs only → use tangent. tan θ = 9/40 = 0.225.
3.7, Two triangles, same sin θ
Triangle A: sin θ = 6/10 = 3/5 = 0.6.
Triangle B: sin θ = 15/25 = 3/5 = 0.6. ✓ identical.
Triangle B is just 2.5× triangle A, same angles → same trig ratios. The scale factor cancels in the ratio opp/hyp.
3.8, Find cos θ when opp = 9, hyp = 15
Pythagoras: adj² = 15² − 9² = 225 − 81 = 144, so adj = 12.
cos θ = adj/hyp = 12/15 = 4/5 = 0.8.
This is a 9–12–15 triangle, which is 3× the 3–4–5. Same angles, same cos.