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).