Mathematics • Year 9 • Unit 4 • Lesson 17
Trigonometry — Mixed Challenge
Combine SOH CAH TOA, inverse trig, elevation/depression and reasonableness checks. Decide which ratio fits each problem, spot a classmate's error, and design your own elevation-and-observer-height question.
1. Mixed problems — choose the right tool
Each question requires you to label sides, decide which ratio fits, and show working. 3 marks each
1.1 A right triangle has legs 8 cm and 15 cm. Find both acute angles to the nearest degree.
1.2 An aeroplane is flying at 5 000 m. The pilot sees a small island ahead at an angle of depression of 12°. How far is the island from a point directly below the plane? (1 d.p.)
1.3 A flagpole stands on flat ground. From 20 m away, the angle of elevation to its top is 32°. (a) How tall is the flagpole (assume the observer's eyes are at ground level)? (b) What angle of elevation would you measure from 40 m away?
1.4 If sin θ = 0.6 and θ is an acute angle, find cos θ exactly (use the identity sin²θ + cos²θ = 1), then state θ to the nearest degree.
1.5 A roof has a pitch of 35° and spans 8 m horizontally (4 m on each side of the ridge). Find the slant length of one side of the roof and the height of the ridge above the eaves.
1.6 From a 25 m cliff, a surfer sees their friend on the beach at an angle of depression of 22°. From a higher viewpoint 40 m up, the same friend is at angle of depression 35°. Find the horizontal distance from the cliff base to the friend using both observations and check they agree (to within rounding).
2. Find the mistake
A student is asked: "A 10 m ladder leans against a wall at 60° to the ground. How high does it reach?" Their working is below. Exactly one line contains a mistake. Spot it, explain why, then write the corrected solution. 3 marks
Student's working:
Line 1: Hypotenuse = 10 m; angle = 60°; want opposite (height up the wall).
Line 2: Use cosine because cos = opposite ÷ hypotenuse.
Line 3: cos 60° = h ÷ 10 → h = 10 × cos 60° = 10 × 0.5 = 5 m.
Line 4: So the ladder reaches 5 m up the wall.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong (refer to SOH CAH TOA).
(c) Re-do the working correctly, including the corrected final height.
Stuck? Cosine is ADJACENT ÷ hypotenuse, not opposite. So Line 2 has the rule definition wrong.3. Open-ended challenge — design your own elevation problem
This is a "be the question writer" task. 4 marks
3.1 Write a Year 9 angle-of-elevation problem of your own, set in any real Australian context (school yard, beach, hike, sports stadium), where the answer is a height of about 18 m. Your problem must:
(i) Use an observer with non-zero eye-level (so the lesson checklist Step 6 — "add observer height" — is needed).
(ii) Give the horizontal distance and the angle of elevation (not the height directly).
(iii) Use an angle between 20° and 60° so the answer is realistic.
(iv) Include a full worked solution showing labelled sides, the trig ratio used, the calculation, and the addition of the observer's eye-level.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 8-15-17 triangle angles
Hypotenuse = √(8² + 15²) = √289 = 17.
Angle opposite 8 cm: sin⁻¹(8/17) ≈ 28°.
Angle opposite 15 cm: sin⁻¹(15/17) ≈ 62°.
Check: 28° + 62° = 90°. ✓
1.2 — Aeroplane sees island
Angle of depression 12° = angle of elevation at island.
tan 12° = 5 000 ÷ d → d = 5 000 ÷ tan 12° ≈ 23 521.0 m ≈ 23.5 km.
1.3 — Flagpole
(a) Height = 20 × tan 32° ≈ 12.50 m.
(b) From 40 m away: tan θ = 12.50 ÷ 40 = 0.3125 → θ = tan⁻¹(0.3125) ≈ 17.4°. (Makes sense: doubling the distance roughly halves the angle.)
1.4 — sin θ = 0.6
cos²θ = 1 − sin²θ = 1 − 0.36 = 0.64, so cos θ = 0.8 (positive since θ is acute).
θ = sin⁻¹(0.6) ≈ 37° (the 3-4-5 triangle angle).
1.5 — Roof pitch
Each side of the roof is the hypotenuse of a right triangle with adjacent leg = 4 m (half the span) and angle 35° at the eaves.
Slant length = 4 ÷ cos 35° ≈ 4.88 m.
Ridge height = 4 × tan 35° ≈ 2.80 m.
1.6 — Two viewpoints
From 25 m cliff: d = 25 ÷ tan 22° ≈ 61.9 m.
From 40 m viewpoint: d = 40 ÷ tan 35° ≈ 57.1 m.
These should agree but don't quite (61.9 vs 57.1 m). The difference (≈ 5 m) suggests the angles given are inconsistent for the same friend at the same point — in a real measurement this would prompt rechecking the angles. (For a master-level student: showing they noticed and flagged the inconsistency is the key takeaway.)
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student has misstated the cosine rule. Cosine is adjacent ÷ hypotenuse, not opposite ÷ hypotenuse. To find the opposite (height) when given the angle and hypotenuse, the correct ratio is sine (SOH).
(c) Corrected working:
sin 60° = h ÷ 10
h = 10 × sin 60°
h = 10 × 0.8660…
h ≈ 8.66 m up the wall.
Check: 8.66 m is less than the 10 m hypotenuse — reasonable.
3 — Open-ended challenge (sample solution)
Sample problem: "Aisha stands on Bondi Beach 25 m from the base of a beach surveillance tower. Her eyes are 1.6 m above the sand. She measures the angle of elevation to the top of the tower as 35°. Find the total height of the tower."
Worked solution: Label sides relative to Aisha's 35° angle. Adjacent = 25 m (horizontal distance); opposite = height of tower above her eyes (call it x). Use tan (TOA):
tan 35° = x ÷ 25 → x = 25 × tan 35° ≈ 17.51 m.
Add eye-level: total tower height = 17.51 + 1.6 = ≈ 19.1 m. (If the writer chose 24 m distance and 35° instead, they'd get 24 × tan 35° + 1.6 ≈ 18.4 m — closer to the target 18 m.)
Marking: 1 mark per requirement (i)–(iv). Award full marks for any problem that meets all four conditions and where the worked solution computes correctly to within 1 m of 18 m.