Mathematics • Year 9 • Unit 3 • Lesson 12
Both Acute Angles — Mixed Practice
Build fluency with finding BOTH acute angles in a right triangle. Use one inverse-trig call, then the complementary-angle rule θ + φ = 90° to get the other angle for free. Walk from one worked example through one guided fill-in to eight independent practice problems.
1. I do — fully worked example
Read every step. Note how we only need ONE inverse-trig calculation: the complementary-angle rule gives us the other acute angle by simple subtraction.
Problem. Find both acute angles of a 3-4-5 right triangle (1 d.p.).
Step 1 — Sketch the triangle and choose a reference angle.
Sides 3, 4, 5 with hypotenuse 5. Call θ the angle opposite the side of length 3.
Reason: labelling helps you pick the right ratio.
Step 2 — Pick a ratio for θ.
Relative to θ: opp = 3, hyp = 5. Use sin (SOH).
Reason: opp + hyp pair → sin.
Step 3 — Apply inverse trig.
sin θ = 3 / 5 = 0.6. θ = sin⁻¹(0.6) ≈ 36.9°
Reason: SHIFT → sin → 0.6 → =.
Step 4 — Use the complementary-angle rule for the second acute angle.
φ = 90° − θ ≈ 90° − 36.9° = 53.1°
Reason: in any right triangle, the two acute angles add to 90° (because the third angle is 90° and all three sum to 180°). One subtraction is faster than another inverse trig call.
Step 5 — Check.
36.9° + 53.1° = 90.0° ✓
Answer: the two acute angles are ≈ 36.9° and ≈ 53.1°.
2. We do — fill in the missing steps
Same structure as Section 1, but the working is faded. Fill in each blank. 4 marks
Problem. Find both acute angles of a 5-12-13 right triangle (1 d.p.).
Step 1 — Sketch: a right triangle with sides 5, 12, 13. The hypotenuse is the side of length ____.
Step 2 — Let θ be the angle opposite the side of length 5. Pick a ratio: relative to θ, opp = 5 and hyp = 13, so use ________ (SOH/CAH/TOA).
Step 3 — Inverse trig:
sin θ = ____ / ____ ≈ 0.____
θ = sin⁻¹( ____ ) ≈ ________ °
Step 4 — Complementary angle:
φ = 90° − ________ ° = ________ °
Step 5 — Check: θ + φ ≈ ________ ° (should be 90°).
3. You do — independent practice
Show working under each problem. Problems 3.1–3.4 are foundation (one acute angle given — find the other). Problems 3.5–3.6 are standard (find both acute angles from two sides). Problems 3.7–3.8 are extension (Pythagoras first, then both angles).
Foundation — apply the complementary-angle rule
3.1 One acute angle of a right triangle is 25°. Find the other. 1 mark
3.2 One acute angle is 42°. Find the other. 1 mark
3.3 One acute angle is 60°. Find the other. 1 mark
3.4 One acute angle is 18.4°. Find the other (1 d.p.). 1 mark
Standard — find BOTH angles from two sides
3.5 A right triangle has opp = 7 and adj = 24 relative to θ. Find both acute angles (1 d.p.). 2 marks
3.6 A right triangle has opp = 9 and hyp = 15 relative to θ. Find both acute angles (1 d.p.). 2 marks
Extension — Pythagoras + both angles
3.7 A right triangle has hyp = 17 cm and one leg = 8 cm. (a) Use Pythagoras to find the other leg. (b) Find both acute angles (1 d.p.). (c) Verify they sum to 90°. 3 marks
3.8 An isosceles right triangle has both legs of length 4 cm. (a) Find both acute angles. (b) Use Pythagoras to find the hypotenuse (leave in exact surd form). 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (5-12-13)
Step 1: hypotenuse is the side of length 13 (it's the longest).
Step 2: use sin (SOH — opp + hyp).
Step 3: sin θ = 5 / 13 ≈ 0.3846. θ = sin⁻¹(5/13) ≈ 22.6°.
Step 4: φ = 90° − 22.6° = 67.4°.
Step 5: θ + φ ≈ 90.0° ✓.
3.1 — One angle 25°
Other = 90° − 25° = 65°.
3.2 — One angle 42°
Other = 90° − 42° = 48°.
3.3 — One angle 60°
Other = 90° − 60° = 30°. (This is the famous 30-60-90 triangle.)
3.4 — One angle 18.4°
Other = 90° − 18.4° = 71.6°.
3.5 — opp = 7, adj = 24
TOA: tan θ = 7 / 24, so θ = tan⁻¹(7/24) ≈ 16.3°.
Other: φ = 90° − 16.3° ≈ 73.7°.
This is the 7-24-25 triangle — check: hyp = √(49 + 576) = √625 = 25.
3.6 — opp = 9, hyp = 15
SOH: sin θ = 9 / 15 = 0.6, so θ = sin⁻¹(0.6) ≈ 36.9°.
Other: φ = 90° − 36.9° ≈ 53.1°.
This is the 9-12-15 triangle (scaled 3-4-5). Same acute angles as the 3-4-5 triangle.
3.7 — hyp = 17, one leg = 8
(a) Other leg = √(17² − 8²) = √(289 − 64) = √225 = 15 cm. (Famous 8-15-17 Pythagorean triple.)
(b) Angle opposite the 8: sin⁻¹(8/17) ≈ 28.1°. Angle opposite the 15: 90° − 28.1° = 61.9°.
(c) 28.1° + 61.9° = 90.0° ✓.
3.8 — Isosceles right triangle, legs = 4
(a) Equal legs → tan θ = 4 / 4 = 1, so θ = tan⁻¹(1) = 45°. By symmetry (and the complementary rule) the other acute angle is also 45°.
(b) Pythagoras: hyp = √(4² + 4²) = √32 = √(16 × 2) = 4√2 cm (exact surd form, ≈ 5.66 cm).