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. ✓