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.

Build · I Do / We Do / You Do

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.

Stuck? Revisit lesson § "The Big Idea" — the three ratios are just the three ways to pair two of the three sides.

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? ____

Stuck? Revisit lesson § "Watch Me Solve It · Compute all three ratios" for the 3-4-5 worked example.

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

Stuck on 3.8? adj² + opp² = hyp². So adj² = 15² − 9² = 225 − 81 = 144, giving adj = 12.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

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.