Mathematics • Year 9 • Unit 4 • Lesson 17
Choosing the Right Trig Ratio
Build fluency with SOH CAH TOA and the inverse trig functions. Practise the lesson's checklist — label sides, identify known and wanted, pick the ratio, solve, check. One fully worked example, one faded example, then eight graduated problems.
1. I do — fully worked example
Notice how the lesson's "label sides → know vs want → pick ratio" routine drives every step.
Problem. A 10 m ladder leans against a wall at an angle of 60° to the ground. How high up the wall does the ladder reach? Give the answer to 2 decimal places.
Step 1 — Label the sides relative to the 60° angle.
Hypotenuse = the 10 m ladder (longest, opposite the right angle at the base).
Opposite the 60° angle = the unknown height up the wall (call it h).
Reason: the side OPPOSITE the angle is the one you can't touch when standing at the angle's vertex.
Step 2 — Know vs want, pick the ratio.
Know: angle (60°), hypotenuse (10 m). Want: opposite (h).
Opposite + Hypotenuse → SINE (SOH).
Reason: sin θ = opposite ÷ hypotenuse.
Step 3 — Substitute and solve.
sin 60° = h ÷ 10
h = 10 × sin 60°
h = 10 × 0.8660…
h ≈ 8.66 m
Step 4 — Reasonableness check.
8.66 m is shorter than the 10 m hypotenuse (good — opposite must always be ≤ hypotenuse) and is positive (good — it's a height). ✓
Answer: the ladder reaches 8.66 m up the wall.
2. We do — fill in the missing steps
Use the same four-step routine. Fill each blank. 4 marks
Problem. From a window 12 m up a building, the angle of depression to a parked car on the ground is 30°. How far is the car from the foot of the building? (Round to 1 d.p.)
Step 1 — Draw a diagram in your head. The 30° angle of depression at the window equals the 30° angle of __________________ at the car (alternate angles).
Step 2 — Label sides relative to the 30° angle at the car.
Opposite the 30° = ____ m (the height of the window).
Adjacent the 30° = d (the unknown ground distance).
Step 3 — Know opposite and want adjacent → use ____________ (T___).
tan 30° = ____ ÷ d
d = ____ ÷ tan 30°
d ≈ ______ m
Step 4 — Check: the answer should be a distance bigger than the 12 m height (because the angle is less than 45°). Is it? ____
3. You do — independent practice
Foundation = one ratio, given the angle and one side. Standard = include observer height or find an angle. Extension = multi-step or angle-of-depression.
Foundation — single ratio
3.1 Find x in a right triangle where x is opposite a 40° angle and the hypotenuse is 15 cm. Give answer to 2 d.p. 1 mark
3.2 Find x in a right triangle where x is adjacent to a 55° angle and the hypotenuse is 20 cm. Give answer to 2 d.p. 1 mark
3.3 A 15 m ladder leans against a wall at 70° to the ground. How high up the wall does it reach? 1 mark
3.4 In a right triangle the side opposite an angle θ is 3 cm and the hypotenuse is 5 cm. Find θ to the nearest degree using sin⁻¹. 1 mark
Standard — combine two ideas
3.5 A ramp rises 1.5 m over a horizontal distance of 12 m. Find the angle of inclination to 1 d.p. 2 marks
3.6 A kite string is 60 m long and makes an angle of 50° with the horizontal ground. Assuming the string is straight and starts at ground level, how high is the kite? 2 marks
Extension — push your thinking
3.7 An observer with eye-level 1.5 m is 40 m from the base of a tree. The angle of elevation to the top of the tree is 35°. Find the total height of the tree to 1 d.p. (Don't forget to add the observer's eye-level — see the lesson checklist.) 3 marks
3.8 From a 45 m high cliff, the angle of depression to a small boat is 18°. How far is the boat from the base of the cliff? Round to 1 d.p. Then verify your answer using a second trig ratio. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded 12 m window / 30°)
Step 1: elevation.
Step 2: opposite = 12 m; adjacent = d.
Step 3: use tangent (TOA). tan 30° = 12 ÷ d → d = 12 ÷ tan 30° ≈ 20.8 m.
Step 4: yes — 20.8 m > 12 m, consistent with the 30° angle being less than 45°.
3.1 — x opposite 40°, hyp 15
sin 40° = x ÷ 15 → x = 15 × sin 40° ≈ 9.64 cm.
3.2 — x adjacent 55°, hyp 20
cos 55° = x ÷ 20 → x = 20 × cos 55° ≈ 11.47 cm.
3.3 — Ladder at 70°
Height = 15 × sin 70° ≈ 14.10 m.
3.4 — sin⁻¹(3/5)
θ = sin⁻¹(3 ÷ 5) = sin⁻¹(0.6) ≈ 37° (the classic 3-4-5 triangle).
3.5 — Ramp angle
tan θ = 1.5 ÷ 12 = 0.125 → θ = tan⁻¹(0.125) ≈ 7.1°.
3.6 — Kite height
Height = 60 × sin 50° ≈ 45.96 m.
3.7 — Tree height with observer
Triangle height (above eye level) = 40 × tan 35° ≈ 28.01 m.
Total tree height = 28.01 + 1.5 = 29.5 m (to 1 d.p.).
The 1.5 m is added because the angle was measured from eye height, not the ground.
3.8 — Cliff to boat
Angle of depression 18° = angle of elevation 18° at the boat (alternate angles).
tan 18° = 45 ÷ d → d = 45 ÷ tan 18° ≈ 138.5 m.
Verification: hypotenuse (line of sight) = 45 ÷ sin 18° ≈ 145.6 m. Check Pythagoras: √(45² + 138.5²) ≈ √(2025 + 19 182) ≈ √21 207 ≈ 145.6 m. ✓