Mathematics • Year 9 • Unit 4 • Lesson 1
Introduction to Trigonometry
Build fluency labelling the opposite, adjacent and hypotenuse of a right-angled triangle, then set up and evaluate sin, cos and tan ratios. One step at a time, from a fully worked example through guided practice to independent problems.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. A right-angled triangle has angle θ = 30°, hypotenuse 12 cm. Find the side opposite θ.
Step 1 — Label the sides relative to the 30° angle.
Hypotenuse = 12 cm (opposite the right angle, always the longest).
Opposite = the side across from 30° — call it x (unknown).
Adjacent = the side next to 30° (not the hypotenuse).
Reason: every trig problem starts by naming sides relative to the angle you're working with.
Step 2 — Pick the right ratio.
We have hypotenuse, we want opposite. SOH → sin = opp/hyp.
Reason: the S in SOH CAH TOA tells you sin connects opposite and hypotenuse.
Step 3 — Set up the equation.
sin 30° = x / 12
Reason: substitute the angle on the left and the side ratio on the right.
Step 4 — Rearrange to make x the subject.
x = 12 × sin 30°
Reason: multiply both sides by 12 to undo the division.
Step 5 — Evaluate using sin 30° = 0.5.
x = 12 × 0.5 = 6 cm
Reason: sin 30° is one of the exact values worth memorising.
Answer: the opposite side is 6 cm.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. A right-angled triangle has angle θ = 45°, hypotenuse 10 cm. Find the adjacent side.
Step 1 — Label sides relative to 45°: hypotenuse = 10 cm, adjacent = ____, opposite = ____.
Step 2 — Pick the ratio. We have hypotenuse, we want adjacent. CAH → cos = ________ / ________ .
Step 3 — Set up:
cos 45° = ____ / 10
Step 4 — Rearrange:
adjacent = 10 × cos 45° = 10 × _______ ≈ _______ cm (2 dp)
Step 5 — State the answer:
adjacent ≈ _________ cm
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (label or single-ratio). The middle two are standard (rearrange to solve). The last two are extension (inverse trig + reasoning).
Foundation — label & evaluate
3.1 A right-angled triangle has angle θ = 25°. Name the three sides relative to θ. (No calculation needed.) 1 mark
3.2 Use your calculator (degree mode) to evaluate sin 60° to 3 decimal places. 1 mark
3.3 Evaluate cos 30° and tan 60° to 3 decimal places. 1 mark
3.4 A triangle has opposite = 5, hypotenuse = 10. Write down (don't solve yet) the equation for sin θ. 1 mark
Standard — solve for a side
3.5 A right-angled triangle has angle 35° and hypotenuse 20 cm. Find the side opposite the 35° angle, to 2 dp. 2 marks
3.6 A right-angled triangle has angle 50° and adjacent side 7 cm. Find the opposite side using tan, to 2 dp. 2 marks
Extension — push your thinking
3.7 If sin θ = 0.766, find θ to the nearest degree. Show the inverse-trig step. 2 marks
3.8 Explain in one or two sentences why sin θ can never be greater than 1 for an angle in a right-angled triangle. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (cos 45°, hyp 10)
Step 1: hypotenuse = 10, adjacent = unknown (next to 45°), opposite = the other unknown side.
Step 2: cos = adjacent / hypotenuse.
Step 3: cos 45° = adj / 10.
Step 4: adjacent = 10 × cos 45° = 10 × 0.7071… ≈ 7.07 cm.
Step 5: adjacent ≈ 7.07 cm.
3.1 — Label sides relative to 25°
Hypotenuse: opposite the right angle (longest side). Opposite: across from the 25° angle. Adjacent: next to the 25° angle, not the hypotenuse.
3.2 — sin 60°
sin 60° ≈ 0.866 (exact value √3 / 2).
3.3 — cos 30° and tan 60°
cos 30° ≈ 0.866 (same as sin 60° — complementary angles).
tan 60° ≈ 1.732 (exact value √3).
3.4 — Equation for sin θ
sin θ = opp / hyp = 5 / 10 = 0.5. (If asked to solve: θ = sin⁻¹(0.5) = 30°.)
3.5 — Opposite side, angle 35°, hyp 20 cm
sin 35° = opp / 20 → opp = 20 × sin 35° = 20 × 0.5736 ≈ 11.47 cm.
3.6 — Opposite side, angle 50°, adj 7 cm
tan 50° = opp / 7 → opp = 7 × tan 50° = 7 × 1.1918 ≈ 8.34 cm.
3.7 — sin θ = 0.766
θ = sin⁻¹(0.766) ≈ 50° (using the inverse sine function on a calculator).
3.8 — Why sin θ ≤ 1
sin θ = opposite / hypotenuse. In any right-angled triangle the hypotenuse is the longest side, so the opposite side is always shorter than or equal to the hypotenuse. The ratio of a shorter length to a longer length is at most 1, so sin θ can never exceed 1.