Mathematics • Year 9 • Unit 3 • Lesson 12
All Angles — Mixed Challenge
Pull together the complementary-angle rule (θ + φ = 90°), inverse trig, Pythagoras, exact-value angles, and the sin = cos-of-complement identity. Choose the cleanest tool for each problem, spot a common Year 9 mistake, and tackle an open-ended challenge about exact-value triangles.
1. Mixed problems — choose the right rule
Each question uses a different combination of complementary angles, Pythagoras, and inverse trig. Decide which tool applies before you start. Show your working. 3 marks each
1.1 A right triangle has opp = 9 cm and adj = 12 cm relative to θ. (a) Find θ to 1 d.p. (b) Find the other acute angle by subtraction. (c) Confirm using a different inverse-trig calculation.
1.2 A right triangle has hypotenuse 25 and one leg 7. Find the other leg (Pythagoras) and BOTH acute angles to 1 d.p.
1.3 Use the complementary-angle identity sin θ = cos(90° − θ) to evaluate sin 25° in two ways and check they agree (1 d.p.).
1.4 Without a calculator, find both acute angles of a right triangle with legs of length 1 cm and √3 cm. (Hint: tan θ = 1 / √3 corresponds to one of the exact-value angles from the standard tables.)
1.5 In a right triangle, one acute angle is twice the other. Find both acute angles exactly. (Hint: let θ + φ = 90° with φ = 2θ.)
1.6 In a right triangle, sin θ = 0.6 and cos θ = 0.8 (same θ). (a) Find θ to 1 d.p. (b) Find the other acute angle φ and confirm that sin φ = 0.8 and cos φ = 0.6 (this is the complementary-angle identity in action).
2. Find the mistake
A Year 9 student is asked to find the second acute angle of a right triangle, given the first acute angle is 36.9°. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — find the other acute angle:
Line 1: The two acute angles of a right triangle add to 180°.
Line 2: So φ = 180° − 36.9° = 143.1°.
Line 3: Check: 36.9 + 143.1 = 180. ✓
Line 4: So the other acute angle is 143.1°.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the correct value of the second acute angle.
Stuck? All three angles of any triangle sum to 180°, but ONE of those three is the 90° right angle — leaving 90° for the TWO acute angles. Also: a 143° angle would be OBTUSE, not acute.3. Open-ended challenge — design a right triangle
This question has more than one valid answer. 4 marks
3.1 Find two different right triangles with acute angles approximately 35° and 55° (each to 1 d.p.). For each triangle:
(i) State its three side lengths (can be decimals or surds — no need to be integers).
(ii) Use Pythagoras to verify the sides form a right triangle.
(iii) Use inverse trig to confirm at least one of the acute angles is ≈ 35°.
(iv) State the other acute angle using the complementary-angle rule (no extra inverse trig).
Bonus: Your two triangles must NOT be similar to each other (they must not have the same side-length ratios).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — opp = 9, adj = 12
(a) TOA: tan θ = 9 / 12 = 0.75. θ = tan⁻¹(0.75) ≈ 36.9°.
(b) Other acute angle = 90° − 36.9° = 53.1°.
(c) Confirm: hyp = √(9² + 12²) = √225 = 15. sin⁻¹(12/15) = sin⁻¹(0.8) ≈ 53.1° ✓.
9-12-15 is just 3-4-5 scaled by 3.
1.2 — hyp = 25, leg = 7
Other leg = √(25² − 7²) = √(625 − 49) = √576 = 24. (Famous 7-24-25 Pythagorean triple.)
Angle opposite 7: sin⁻¹(7/25) ≈ 16.3°.
Other acute angle: 90° − 16.3° ≈ 73.7°.
1.3 — sin 25° via complementary identity
Direct: sin 25° ≈ 0.4 (1 d.p.).
Via identity: sin 25° = cos(90° − 25°) = cos 65° ≈ 0.4 (1 d.p.) ✓.
Both give the same value, confirming sin θ = cos(90° − θ).
1.4 — Legs 1 and √3
Take θ as the angle opposite the leg of length 1: tan θ = 1 / √3. From the exact-value table, tan 30° = 1/√3, so θ = 30° exactly.
Other acute angle = 90° − 30° = 60°.
This is the classic 30-60-90 triangle (the half of an equilateral triangle).
1.5 — One angle twice the other
Let θ be the smaller. Then φ = 2θ and θ + φ = 90°, so θ + 2θ = 90°, giving 3θ = 90° and θ = 30°. The other acute angle is φ = 2 × 30° = 60°. (This is the 30-60-90 triangle again, derived from the angle ratio.)
1.6 — sin θ = 0.6 and cos θ = 0.8
(a) θ = sin⁻¹(0.6) ≈ 36.9°. (Quick check: cos 36.9° ≈ 0.8 ✓.)
(b) φ = 90° − 36.9° ≈ 53.1°. sin 53.1° ≈ 0.8 ✓ and cos 53.1° ≈ 0.6 ✓.
The pattern sin θ = cos φ and cos θ = sin φ when θ + φ = 90° is the complementary-angle identity in action.
2 — Find the mistake
(a) The mistake is on Line 1 (which then makes Line 2 and Line 4 wrong too).
(b) ALL THREE angles of any triangle sum to 180°, but one of those three IS the right angle (90°). That leaves only 90° to share between the TWO acute angles. The two ACUTE angles sum to 90°, not 180°. Also, an angle of 143.1° is obtuse, not acute — that should have been a red flag.
(c) Corrected working:
The two acute angles of a right triangle add to 90° (the third angle is 90°, and all three angles of any triangle sum to 180°).
So φ = 90° − 36.9° = 53.1°.
Check: 36.9° + 53.1° = 90.0° ✓.
3 — Open-ended challenge (sample solution)
For 35°, tan 35° ≈ 0.7002. So I need opp / adj ≈ 0.7 relative to the 35° angle.
Triangle 1: opp = 7, adj = 10. Then hyp = √(7² + 10²) = √149 ≈ 12.21.
Check Pythagoras: 7² + 10² = 49 + 100 = 149 = (√149)² ✓.
Inverse trig: tan⁻¹(7/10) = tan⁻¹(0.7) ≈ 35.0° ✓.
Other acute angle: 90° − 35.0° = 55.0° (by complementary rule).
Triangle 2: opp = 14, adj = 22, hyp = √(14² + 22²) = √(196 + 484) = √680 ≈ 26.08.
Check: tan⁻¹(14/22) = tan⁻¹(0.6364) ≈ 32.5°. (Too small.) Better try opp = 5, adj = 7.14 → tan = 0.700 → θ ≈ 35.0° ✓. Then hyp = √(5² + 7.14²) ≈ √76.0 ≈ 8.72. Ratio is 5 : 7.14 : 8.72 — not the same as 7 : 10 : 12.21, so the two triangles are NOT similar.
Other acute angle: 90° − 35.0° = 55.0°.
Other valid approaches: any pair of triangles with tan ≈ 0.7002 and DIFFERENT scale factors works. Accept any two triangles whose Pythagoras checks out and whose inverse trig confirms ≈ 35°.
Marking: 2 marks per valid triangle (1 for sides + Pythagoras check, 1 for inverse-trig + complementary-rule check). Up to 4 in total. Strictly non-similar pair earns full marks; similar pairs get 3/4.