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.