Mathematics Advanced • Year 11 • Module 2 • Lesson 5
Exact Values & Special Triangles
Apply exact values to geometry, surveying, pendulum motion, and identity-verification problems — without ever touching a calculator.
Problem 1 — Square diagonal (the lesson's hook)
A square has side length 2 cm. A diagonal is drawn, forming two 45-45-90 triangles.
Set up: What are we solving for?
(i) Find the exact length of the diagonal using the 45-45-90 side ratio 1 : 1 : √2. 2 marks
(ii) Verify the same answer using the trigonometric formula: diagonal = side / cos 45°. (Set up the right triangle with the diagonal as hypotenuse, one side adjacent to the 45° angle.) 2 marks
(iii) A second square has area 32 cm². Find the exact length of its diagonal. 3 marks
Stuck on (iii)? Side length = √area = √32 = 4√2. Then apply the 1:1:√2 ratio (or the trig method from (ii)).Problem 2 — Equilateral triangle height (geometric)
An equilateral triangle has side length 6 cm. The altitude from one vertex to the opposite side splits the triangle into two congruent 30-60-90 triangles.
Set up: What are we solving for?
(i) Identify the three sides of one half-triangle: the hypotenuse (= one side of the equilateral), the base (= half a side), and the altitude (the unknown). Find the altitude using exact trig values — explicitly which ratio links a 60° angle to opposite (altitude) and hypotenuse (6 cm). 3 marks
(ii) Find the exact area of the equilateral triangle using A = ½ × base × height with your altitude from (i). 2 marks
(iii) A general formula: show algebraically that an equilateral triangle of side s has area s²√3/4. Use the same trig setup with side s instead of 6. 2 marks
Problem 3 — Surveying tower height (data)
A surveyor records the angle of elevation to the top of a communications tower from three different distances (all on level ground). She wants exact-value answers wherever possible.
| Position | Distance to base (m) | Angle of elevation | Tower height (exact) |
|---|---|---|---|
| A | 30 | 60° | |
| B | 30√3 | 30° | |
| C | 30 | 45° |
Set up: What are we solving for?
(i) Fill in the "tower height (exact)" column. Use tan(angle) = height / distance for each row. 3 marks
(ii) Two rows give the same tower height. Which two, and what is that height? 2 marks
(iii) The surveyor expects all three readings to give the same height (they're measuring the same tower). One reading is inconsistent. Which row is the outlier, and by what factor does it differ from the consistent height? 2 marks
Stuck on (iii)? Compute all three heights as exact expressions, then compare.Problem 4 — Pendulum displacement (modelling)
A pendulum of length 0.8 m swings from the vertical to a maximum angle of θ on each side. At maximum displacement, the bob is at a horizontal distance x = 0.8 sin θ from the rest position and a height h = 0.8 (1 − cos θ) above its lowest point.
Set up: What are we solving for?
(i) When θ = 30°, find the exact horizontal displacement x and the exact height h above the rest position. 2 marks
(ii) When θ = 60°, find x and h in exact form. Compare with the θ = 30° values — is the bob twice as high? Why or why not? 3 marks
(iii) Show that for a small swing θ = 45°, the bob's height h = 0.8(1 − √2/2) m. Then simplify to a single fraction with a rationalised denominator. 2 marks
Problem 5 — Identity verification
A famous identity (which you will meet formally in Year 12) is the double-angle formula: sin(2θ) = 2 sin θ cos θ. Here we verify it using exact values for several special angles.
Set up: What are we solving for?
(i) For θ = 30°: compute 2 sin 30° cos 30° in exact form, then compare with sin 60°. 2 marks
(ii) For θ = 45°: compute 2 sin 45° cos 45°, then compare with sin 90°. 2 marks
(iii) For θ = 60°: compute 2 sin 60° cos 60°, then compare with sin 120°. Does the identity hold in all three cases? 2 marks
Stuck? Substitute the exact value, multiply out, and simplify. The Pythagorean identity may help in (ii).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Square diagonal
Set up. A square's diagonal is the hypotenuse of an isosceles right triangle whose legs are the square's sides. We use the 45-45-90 ratio 1:1:√2 or the trig identity cos 45° = adj/hyp.
(i) Side ratio 1:1:√2 scaled by side = 2: diagonal = 2√2 cm. ✓ 2√2 cm (≈ 2.83 cm).
(ii) cos 45° = side/diagonal = 2/d ⇒ d = 2/cos 45° = 2/(√2/2) = 4/√2 = 4√2/2 = 2√2 cm. ✓
(iii) Side = √32 = √(16 × 2) = 4√2 cm. Diagonal = side × √2 = 4√2 × √2 = 4 × 2 = 8 cm.
Problem 2 — Equilateral triangle height
Set up. Splitting the equilateral triangle in half gives a 30-60-90 right triangle: hypotenuse = 6 cm (full side), base = 3 cm (half side), altitude = unknown opposite the 60° angle.
(i) sin 60° = opp/hyp = h/6 ⇒ h = 6 × sin 60° = 6 × √3/2 = 3√3 cm. The ratio used is sine (opposite over hypotenuse).
(ii) Area = ½ × base × height = ½ × 6 × 3√3 = 9√3 cm² (≈ 15.59 cm²).
(iii) Half-triangle for general side s: hyp = s, half-base = s/2, altitude = s sin 60° = s √3/2. Area of full equilateral = ½ × s × s√3/2 = s²√3/4. ✓ (Check with s = 6: 36√3/4 = 9√3 ✓.)
Problem 3 — Surveying tower
Set up. For each position, tan(angle) = height/distance, so height = distance × tan(angle), using exact values.
(i) A: 30 × tan 60° = 30 × √3 = 30√3 m.
B: 30√3 × tan 30° = 30√3 × √3/3 = 30 × 3/3 = 30 m.
C: 30 × tan 45° = 30 × 1 = 30 m.
(ii) B and C both give height = 30 m.
(iii) The outlier is row A (30√3 m ≈ 51.96 m vs the consistent 30 m). It differs by a factor of √3. (Likely the surveyor mis-measured either the distance or the angle at position A.)
Problem 4 — Pendulum displacement
Set up. Length L = 0.8 m. Horizontal displacement x = L sin θ; vertical rise above lowest point h = L(1 − cos θ).
(i) θ = 30°: x = 0.8 sin 30° = 0.8 × 1/2 = 0.4 m. h = 0.8(1 − cos 30°) = 0.8(1 − √3/2) = 0.8 × (2 − √3)/2 = 0.4(2 − √3) m (≈ 0.107 m).
(ii) θ = 60°: x = 0.8 sin 60° = 0.8 × √3/2 = 0.4√3 m (≈ 0.693 m). h = 0.8(1 − cos 60°) = 0.8(1 − 1/2) = 0.8 × 1/2 = 0.4 m. So h at 60° / h at 30° = 0.4 / 0.4(2 − √3) = 1/(2 − √3) ≈ 3.73 — the bob is not twice as high; it is about 3.73 times as high. The height is non-linear in θ because the (1 − cos θ) factor is concave-up for small θ.
(iii) h = 0.8(1 − √2/2) = 0.8 × (2 − √2)/2 = 0.4(2 − √2) m (≈ 0.234 m). All denominators rationalised; already in simplest exact form.
Problem 5 — Identity verification (double-angle)
Set up. We test sin(2θ) = 2 sin θ cos θ at three special angles using exact values.
(i) θ = 30°: 2 sin 30° cos 30° = 2 × 1/2 × √3/2 = √3/2. sin(60°) = √3/2. ✓ Holds.
(ii) θ = 45°: 2 sin 45° cos 45° = 2 × √2/2 × √2/2 = 2 × 2/4 = 1. sin(90°) = 1. ✓ Holds.
(iii) θ = 60°: 2 sin 60° cos 60° = 2 × √3/2 × 1/2 = √3/2. sin(120°) = +sin 60° (QII positive) = √3/2. ✓ Holds.
The identity holds in all three cases, as it must — sin(2θ) = 2 sin θ cos θ is an identity (true for every θ), and exact-value testing is a powerful way to develop confidence before the formal proof in Year 12.