Mathematics • Year 9 • Unit 3 • Lesson 12
Both Acute Angles in the Real World
Use the complementary-angle rule (θ + φ = 90°) plus inverse trig to find BOTH acute angles in everyday right triangles: a roof truss, a skateboard ramp, a kite string, a pitched solar panel, and a ladder. Then explain when one inverse trig call is enough.
1. Word problems
Each problem asks for at least one acute angle. Use the complementary-angle rule whenever it saves work. Show your working — a single final answer with no working only earns half marks. Round angles to 1 d.p. unless stated.
1.1 — Roof truss. A symmetric "A-frame" roof truss spans 12 m and has an apex 3 m above the centre of the span. The roof forms two identical right triangles sharing a vertical centre line.
(a) Sketch ONE of the right triangles (half the roof).
(b) Find the pitch (the angle the roof makes with the horizontal) to 1 d.p.
(c) Find the other acute angle of the right triangle. 3 marks
1.2 — Solar panel pitch. An engineer mounts a solar panel as the hypotenuse of a right triangle. The panel is 2.5 m long; its highest point is 0.8 m above its lowest point.
(a) Find the tilt angle (acute angle to the horizontal) to 1 d.p.
(b) Find the other acute angle (the angle the back support makes with the panel).
(c) For Sydney, the optimal solar tilt is around 33°. Comment on whether this installation is well-suited to Sydney. 3 marks
1.3 — Kite string. Mei flies a kite. Her hand is at the corner of a right triangle: her kite string (the hypotenuse) is 30 m long; the kite is 24 m directly above the ground at the kite's spot. (Treat the string as straight.)
(a) Find the angle Mei's string makes with the horizontal (1 d.p.).
(b) Find the angle the string makes with the vertical (1 d.p.).
(c) Verify both angles sum to 90°. 3 marks
1.4 — Skateboard quarterpipe. A quarterpipe has a vertical wall 1.2 m tall sitting at the back of a wooden base 1.6 m long (horizontal). The ride surface is the hypotenuse from the top of the wall down to the front of the base.
(a) Find the length of the ride surface (Pythagoras).
(b) Find BOTH acute angles of the right triangle (1 d.p.). State which is the "transition" angle (the angle the wall makes with the ride surface). 3 marks
1.5 — Surveyor's check. A surveyor measures the angle between a sightline to a treetop and the horizontal as 38°. Quickly, without using inverse trig again, state the angle the sightline makes with the VERTICAL. Then explain in one sentence why this works for ANY right-triangle setup. 2 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate finds the first acute angle of a 5-12-13 right triangle as ≈ 22.6° using sin⁻¹. They then re-do the inverse trig calculation from scratch for the second angle: cos⁻¹(5/13). In your own words, explain (i) why they're wasting effort, (ii) which faster method should they use, and (iii) what the faster method gives as the second acute angle. Use the phrase "complementary angles" somewhere in your explanation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Roof truss
(a) Half-roof triangle: adj = 6 m (half the 12 m span), opp = 3 m (apex), hyp = √(6² + 3²) = √45 ≈ 6.71 m (the slanted roof line).
(b) Pitch: tan θ = 3 / 6 = 0.5, so θ = tan⁻¹(0.5) ≈ 26.6°.
(c) Other acute angle = 90° − 26.6° = 63.4° (the angle at the apex between the centre line and the roof slope).
1.2 — Solar panel pitch
(a) sin θ = 0.8 / 2.5 = 0.32, so tilt θ = sin⁻¹(0.32) ≈ 18.7°.
(b) Other acute angle = 90° − 18.7° = 71.3°.
(c) 18.7° is well below Sydney's optimal ≈ 33°, so this panel is tilted too gently and will under-perform in winter when the sun is lower; the engineer could raise the back of the panel for better year-round output.
1.3 — Kite string
(a) sin θ = 24 / 30 = 0.8, so angle with horizontal = sin⁻¹(0.8) ≈ 53.1°.
(b) Angle with vertical = 90° − 53.1° = 36.9° (complementary rule — no extra inverse-trig needed).
(c) 53.1° + 36.9° = 90.0° ✓.
Sense check: this is the same 3-4-5 triangle in disguise (3-4-5 scaled by 6 gives 18-24-30 — close enough that the angles match).
1.4 — Skateboard quarterpipe
(a) Ride surface = √(1.2² + 1.6²) = √(1.44 + 2.56) = √4 = 2 m. (1.2-1.6-2 is a scaled 3-4-5.)
(b) Angle at base of wall (between wall and ride surface) = "transition angle". Relative to the angle at the front of the base (between base and ride surface): opp = 1.2 (wall), adj = 1.6 (base). tan = 1.2/1.6 = 0.75, so this angle ≈ 36.9°.
The OTHER acute angle (the transition angle at the top of the wall) = 90° − 36.9° = 53.1°.
1.5 — Surveyor's check
Angle with the vertical = 90° − 38° = 52°. This works because the horizontal and vertical are 90° apart, so the sightline always sits between them; the two angles add to a right angle, exactly the complementary-angle rule from Lesson 12.
2.1 — Explain your thinking (sample response)
My classmate is wasting effort because the two acute angles in any right triangle are complementary — they add to 90°. Once you have one of them, the other is just 90° minus that, so a SECOND inverse-trig calculation is unnecessary. The faster method is to compute the first angle as θ = sin⁻¹(5/13) ≈ 22.6°, then immediately get φ = 90° − 22.6° ≈ 67.4°. This avoids any new SHIFT-trig keystrokes (and stops rounding errors compounding across two separate calculations).
Marking: 1 mark for identifying "complementary" or the sum-to-90° rule; 1 for naming the faster method (90° − θ); 1 for the correct second angle 67.4°; 1 for clear, full-sentence explanation.