Mathematics Advanced • Year 11 • Module 2 • Lesson 4
Trigonometric Ratios
Build procedural fluency in SOH CAH TOA, choosing the correct ratio, and applying the Pythagorean identity to find missing trig values.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete each definition for an acute angle θ in a right-angled triangle:
sin θ = ____________ / ____________
cos θ = ____________ / ____________
tan θ = ____________ / ____________
Q1.2 Write tan θ in terms of sin θ and cos θ: tan θ = ____________ / ____________
Q1.3 Write the Pythagorean identity: ____________ + ____________ = 1.
2. Worked example, finding a missing side
Follow each line of algebra. Every step has a reason on the right.
Problem. A right-angled triangle has a 30° angle, a hypotenuse of 8 cm, and an unknown side opposite the 30° angle. Find the missing side.
Step 1, Label the three sides relative to θ.
Hypotenuse = 8 cm. Opposite = x (the unknown). Adjacent = unknown (not asked).
Reason: SOH CAH TOA, pick the ratio that links what you know to what you want.
Step 2, Choose the correct ratio: opposite + hypotenuse = sine.
sin 30° = opp / hyp = x / 8.
Reason: SOH (sine = opposite over hypotenuse).
Step 3, Substitute the exact value sin 30° = 1/2.
1/2 = x / 8.
Reason: 30° is a special angle, use exact value, not 0.5 from a calculator.
Step 4, Solve.
x = 8 × 1/2 = 4 cm.
Conclusion. The opposite side is 4 cm.
3. Faded example, finding cos from tan
Given tan θ = 3/4 and θ is acute, find the exact value of cos θ. Fill in each blank. 4 marks
Step 1, Interpret tan θ = 3/4 as a right triangle ratio.
tan θ = opp / adj = ________ / ________. So we can take opp = 3, adj = 4.
Step 2, Find the hypotenuse using Pythagoras.
hyp = √(3² + 4²) = √(________ + ________) = √________ = ________.
Step 3, Apply CAH: cos θ = adj / hyp.
cos θ = ________ / ________.
Step 4, State the answer (and note θ is acute, so the answer is positive).
cos θ = ________________ (the (3, 4, 5) Pythagorean triple makes this clean).
Conclusion. cos θ = ________________.
4. Graduated practice, trig ratio fluency
Show one line of substitution and one of simplification for each.
Foundation, recall and substitute (4 questions)
| Q | Problem | Answer (exact form) |
|---|---|---|
| 4.1 1 | sin θ = 3/5, hyp = 20 cm. Find opp. | |
| 4.2 1 | cos θ = 4/5, hyp = 15 cm. Find adj. | |
| 4.3 1 | tan θ = 2, adj = 3 cm. Find opp. | |
| 4.4 1 | sin θ = 1/√2 (acute). Find tan θ. |
Standard, typical HSC difficulty (6 questions)
Use exact values where possible. Choose the correct ratio first.
4.5 A right triangle has hypotenuse 12 cm and an angle of 60°. Find the side opposite the 60° angle (exact form). 2 marks
4.6 A ramp makes a 15° angle with the ground; the horizontal distance is 10 m. Find the ramp length (hypotenuse) to 1 decimal place. 2 marks
4.7 Given cos θ = 5/13 and θ is acute, find sin θ (exact form). 2 marks
4.8 In a right triangle, adj = 6 cm and hyp = 12 cm. Find the angle θ in exact form (degrees). 2 marks
4.9 Given sin θ = 1/2 (acute), find tan θ (exact form). 2 marks
4.10 A right triangle has opp = 9 cm and angle θ with tan θ = 3/2. Find adj and hyp (exact form). 2 marks
Extension, combine concepts (2 questions)
4.11 Given sin θ = 3/5 (no quadrant specified), find both possible values of cos θ and explain which quadrants each corresponds to. 3 marks
4.12 Show that the expression sin²θ + cos²θ + tan² 45° = 2 for any θ. (Hint: use the Pythagorean identity, then substitute the exact value of tan 45°.) 3 marks
5. Self-check the easy 3
Tick the first three once you've verified your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, SOH CAH TOA
sin θ = opposite / hypotenuse. cos θ = adjacent / hypotenuse. tan θ = opposite / adjacent.
Q1.2, tan in terms of sin and cos
tan θ = sin θ / cos θ (follows from the ratio definitions, since the hypotenuses cancel).
Q1.3, Pythagorean identity
sin²θ + cos²θ = 1.
Q3, Faded example: tan θ = 3/4 acute, find cos θ
Step 1: tan θ = opp/adj = 3/4.
Step 2: hyp = √(3² + 4²) = √(9 + 16) = √25 = 5.
Step 3: cos θ = 4/5.
Step 4: cos θ = 4/5.
Conclusion: cos θ = 4/5.
Q4.1, sin θ = 3/5, hyp = 20
opp = sin θ × hyp = 3/5 × 20 = 12 cm.
Q4.2, cos θ = 4/5, hyp = 15
adj = cos θ × hyp = 4/5 × 15 = 12 cm.
Q4.3, tan θ = 2, adj = 3
opp = tan θ × adj = 2 × 3 = 6 cm.
Q4.4, sin θ = 1/√2, acute
1/√2 = √2/2 (rationalised), which is sin 45°. So θ = 45° and tan θ = 1.
Q4.5, hyp = 12, θ = 60°, find opp
sin 60° = opp/12. sin 60° = √3/2, so opp = 12 × √3/2 = 6√3 cm (≈ 10.39 cm).
Q4.6, Ramp: adj = 10, θ = 15°, find hyp
cos 15° = 10/L ⇒ L = 10/cos 15° ≈ 10/0.9659 ≈ 10.4 m.
Q4.7, cos θ = 5/13, acute, find sin θ
Pythagorean: sin²θ = 1 − 25/169 = 144/169, sin θ = ±12/13. Acute ⇒ positive: sin θ = 12/13. (The (5, 12, 13) Pythagorean triple makes this clean.)
Q4.8, adj = 6, hyp = 12
cos θ = 6/12 = 1/2. The exact angle with cos = 1/2 is θ = 60° (equivalently π/3 rad).
Q4.9, sin θ = 1/2 acute, find tan θ
θ = 30°, so tan θ = tan 30° = 1/√3 = √3/3 (rationalised).
Q4.10, opp = 9, tan θ = 3/2
tan θ = opp/adj = 9/adj = 3/2 ⇒ adj = 9 × 2/3 = 6 cm. Then hyp = √(9² + 6²) = √(81 + 36) = √117 = 3√13 cm.
Q4.11, sin θ = 3/5, both quadrants
sin positive ⇒ θ is in QI or QII. cos²θ = 1 − 9/25 = 16/25, so cos θ = ±4/5.
• QI: cos θ = 4/5 (both positive).
• QII: cos θ = −4/5 (sin positive, cos negative).
Q4.12, sin²θ + cos²θ + tan² 45° = 2
By Pythagorean identity, sin²θ + cos²θ = 1 for any θ. Also tan 45° = 1 (from the 45-45-90 special triangle), so tan² 45° = 1. Sum = 1 + 1 = 2. ✓ (Holds for every θ in the domain of sin and cos, i.e. every real number.)