Mathematics • Year 9 • Unit 3 • Lesson 7
Sine Ratio — Finding a Side
Use sin θ = opp/hyp to find the opposite (opp = hyp × sin θ) or the hypotenuse (hyp = opp ÷ sin θ). Build from a fully-worked example through guided practice to eight independent problems. Calculator in DEG mode throughout.
1. I do — fully worked example
A ramp 8 m long rises at 15° to the ground. Find the vertical height (the opp). Each step is reasoned line-by-line.
Problem. hyp = 8 m, θ = 15°. Find opp to 2 d.p. Make sure your calculator is in DEG mode (quick check: sin 30° must give 0.5).
Step 1 — Label sides relative to θ.
hyp = 8 m (the ramp itself — the longest, slanted side), opp = ? (the vertical height across from θ), θ = 15°.
Reason: we have the hyp + want the opp → that side-pair calls for sine (SOH).
Step 2 — Set up the sine equation.
sin 15° = opp / 8
Reason: sin θ = opp/hyp.
Step 3 — Rearrange to isolate opp.
opp = 8 × sin 15°
Reason: multiply both sides of the equation by 8 to move it from the denominator to a multiplier.
Step 4 — Compute (DEG mode).
opp ≈ 8 × 0.2588 ≈ 2.07 m
Reason: keep full precision in the calculator, then round the final answer to 2 d.p.
Step 5 — Add units, sanity check.
opp ≈ 2.07 m. Reasonable? Yes — a shallow 15° ramp of 8 m only rises a couple of metres.
Answer: opp ≈ 2.07 m.
2. We do — fill in the missing steps
A right triangle has an angle of 28° with the opposite side equal to 5 cm. Find the hypotenuse to 2 d.p. Fill in each blank. 4 marks
Step 1 — Label. opp = ____ cm, hyp = ?, θ = ____ ° . The two sides involved are opp and hyp, so the ratio is __________ .
Step 2 — Set up:
sin ____ ° = ____ / hyp
Step 3 — Rearrange to put hyp on its own:
hyp = ____ / sin ____ °
Step 4 — Compute (DEG mode):
hyp ≈ 5 / ________ ≈ ________ cm
Step 5 — Sanity check. Should the hypotenuse be bigger or smaller than the opposite? __________ . Is your answer? __________
3. You do — independent practice
Round to 2 d.p. unless told otherwise. Foundation (3.1–3.4) is a single multiply/divide. Standard (3.5–3.6) needs you to pick a side. Extension (3.7–3.8) is multi-step.
Foundation — opp = hyp × sin θ, or hyp = opp ÷ sin θ
3.1 hyp = 12, θ = 35°. Find opp. 1 mark
3.2 hyp = 25, θ = 30°. Find opp. (Use sin 30° = 0.5 exactly.) 1 mark
3.3 opp = 6, θ = 45°. Find hyp. 1 mark
3.4 opp = 4, θ = 60°. Find hyp. 1 mark
Standard — combine ideas
3.5 A skateboard ramp 1.5 m long meets the ground at 20°. Find the vertical height the ramp reaches. 2 marks
3.6 A kite string is at 55° to the horizontal. The kite is 18 m vertically above the kite-flyer's hand. Find the length of the kite string (the hypotenuse). 2 marks
Extension — push your thinking
3.7 A ramp of length L meets the ground at 25°. The vertical rise is 3.5 m. Find L to 2 d.p. and then find the horizontal length of the ramp (the adj) using Pythagoras. 3 marks
3.8 Two ramps are 6 m long. Ramp A is at 20° to the ground; Ramp B is at 40°. Find the vertical height of each (2 d.p.) and write one sentence explaining what doubling the angle did to the height. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (28°, opp = 5)
Step 1: opp = 5, θ = 28°, ratio is sin.
Step 2: sin 28° = 5 / hyp.
Step 3: hyp = 5 / sin 28°.
Step 4: hyp ≈ 5 / 0.4695 ≈ 10.65 cm.
Step 5: hyp must be BIGGER than opp (it's always the longest side); 10.65 > 5 ✓.
3.1 — hyp = 12, θ = 35°
opp = 12 sin 35° ≈ 12 × 0.5736 ≈ 6.88.
3.2 — hyp = 25, θ = 30°
opp = 25 sin 30° = 25 × 0.5 = 12.5 exactly.
3.3 — opp = 6, θ = 45°
hyp = 6 / sin 45° ≈ 6 / 0.7071 ≈ 8.49.
3.4 — opp = 4, θ = 60°
hyp = 4 / sin 60° ≈ 4 / 0.8660 ≈ 4.62.
3.5 — Skateboard ramp 1.5 m at 20°
opp + hyp involved → sin. height = 1.5 sin 20° ≈ 1.5 × 0.3420 ≈ 0.51 m (about half a metre).
3.6 — Kite string
opp (height) = 18, want hyp (string). sin 55° = 18/hyp, so hyp = 18 / sin 55° ≈ 18 / 0.8192 ≈ 21.97 m.
3.7 — Ramp, rise = 3.5 m, θ = 25°
L = hyp = 3.5 / sin 25° ≈ 3.5 / 0.4226 ≈ 8.28 m.
Then adj² = L² − opp² ≈ 8.28² − 3.5² ≈ 68.56 − 12.25 ≈ 56.31, so adj ≈ 7.50 m.
Sense check: long, shallow ramp — horizontal much longer than vertical, as expected.
3.8 — Two 6 m ramps
Ramp A (20°): height = 6 sin 20° ≈ 6 × 0.3420 ≈ 2.05 m.
Ramp B (40°): height = 6 sin 40° ≈ 6 × 0.6428 ≈ 3.86 m.
Doubling the angle did NOT double the height — the height almost doubled (2.05 → 3.86 is ×1.88), but not exactly, because sin is a non-linear function. The "ratio of heights" is sin 40°/sin 20° ≈ 1.88, not 2.