Mathematics Advanced • Year 11 • Module 2 • Lesson 5
Exact Values & Special Triangles
Practise HSC-style writing on exact values, including the derivation of the 30-60-90 ratios and a structured proof problem.
1. Short-answer questions
1.1 Find the exact value of cos 330°. Show the reference angle and the ASTC sign reasoning. 2 marks Band 3
1.2 Find the exact value of 2 sin 60° cos 60°. Simplify your answer fully. 2 marks Band 3-4
1.3 Without using a calculator, show that:
tan 60° − tan 30° = 2√3/3.
Then state — without further computation — the value of tan 30° − tan 60°. 3 marks Band 4
1.4 Evaluate sin² 30° + cos² 30° + tan 45° in exact form, and explain in one sentence which identity is being applied to the first two terms. 2 marks Band 3-4
Stuck on 1.3? Once you've shown the first claim, the second is simply the negative.2. Extended response
2.1 This question derives the 30-60-90 special triangle and then applies it.
(a) An equilateral triangle has all sides of length 2 and all angles 60°. Drop the altitude from one vertex to the opposite side, splitting the equilateral triangle into two congruent right triangles. Prove, using Pythagoras' theorem, that these right triangles have sides in the ratio 1 : √3 : 2 (shortest leg : longer leg : hypotenuse).
(b) Hence, using only the definitions sin θ = opp/hyp, cos θ = adj/hyp, and tan θ = opp/adj, derive the exact values of sin 30°, cos 30°, tan 30°, sin 60°, cos 60°, and tan 60°.
(c) Using the exact values from (b), evaluate the expression
(sin 60° + cos 30°) × tan 30°
in exact form (with rationalised denominator). 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — identifies the half-triangle's three sides: hypotenuse = 2 (full side), shortest leg = 1 (half-base), altitude unknown.
• 1 mark — applies Pythagoras: altitude² = 2² − 1² = 4 − 1 = 3, so altitude = √3.
• 1 mark — concludes side ratio 1 : √3 : 2, identifying which side is opposite each of the 30°, 60°, 90° angles.
Part (b) — 3 marks
• 1 mark — correctly identifies "opposite" and "adjacent" for the 30° angle in the half-triangle (opp = 1, adj = √3, hyp = 2).
• 1 mark — gives sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3 = √3/3 (rationalised).
• 1 mark — gives sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3, with clear connection to the same triangle (sin 60° = cos 30° via complementarity).
Part (c) — 2 marks
• 1 mark — substitutes and simplifies the bracket: sin 60° + cos 30° = √3/2 + √3/2 = √3.
• 1 mark — multiplies: √3 × √3/3 = 3/3 = 1.
Your response:
Stuck on (a)? Draw the equilateral triangle with the altitude; the altitude bisects the base because the triangle is symmetric.How did this worksheet feel?
What I'll revisit before next class:
1.1 — cos 330° (2 marks)
Sample response. 330° is in Quadrant IV. Reference angle: α = 360° − 330° = 30°. cos 30° = √3/2. In QIV, cosine is positive (ASTC: only cos positive in QIV). Therefore cos 330° = √3/2.
Marking notes. 1 mark — correct reference angle and quadrant. 1 mark — final answer √3/2 with explicit ASTC justification (not just "from the unit circle"). Common error: states −√3/2 by misapplying ASTC (loses 1 mark).
1.2 — 2 sin 60° cos 60° (2 marks)
Sample response. 2 × (√3/2) × (1/2) = 2 × √3/4 = √3/2. (Side note: this is exactly sin 120° — the double-angle identity sin 2θ = 2 sin θ cos θ at work.)
Marking notes. 1 mark — correct substitution of exact values. 1 mark — correct simplified final value √3/2.
1.3 — tan 60° − tan 30° (3 marks)
Sample response. tan 60° = √3 and tan 30° = 1/√3 = √3/3 (rationalised). With common denominator 3:
tan 60° − tan 30° = 3√3/3 − √3/3 = 2√3/3. ✓ (Hence shown.)
Then tan 30° − tan 60° = −(tan 60° − tan 30°) = −2√3/3 (negative of the previous result).
Marking notes. 1 mark — correct substitution and rationalisation. 1 mark — arrives at 2√3/3 via common denominator. 1 mark — correctly identifies the second value as the negative without recomputing.
1.4 — sin² 30° + cos² 30° + tan 45° (2 marks)
Sample response. By the Pythagorean identity, sin² 30° + cos² 30° = 1 (or compute directly: (1/2)² + (√3/2)² = 1/4 + 3/4 = 1). tan 45° = 1. Sum = 1 + 1 = 2.
The identity used on the first two terms is the Pythagorean identity sin²θ + cos²θ = 1.
Marking notes. 1 mark — correct value 2. 1 mark — explicit naming of the Pythagorean identity (or equivalent direct computation).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Deriving the 30-60-90 side ratio. Consider an equilateral triangle with all sides 2 and all angles 60°. Drop the altitude from the apex perpendicular to the base. By symmetry, this altitude bisects the base, splitting the equilateral triangle into two congruent right triangles. Each right triangle has:
• hypotenuse = 2 (a full side of the equilateral triangle),
• shorter leg = 1 (half the base),
• longer leg = altitude = unknown. [1 mark]
By Pythagoras' theorem applied to this right triangle:
altitude² + 1² = 2² ⇒ altitude² = 4 − 1 = 3 ⇒ altitude = √3. [1 mark]
The three angles of the half-triangle are 30° (at the apex, half of 60°), 60° (at the base of the equilateral, unchanged), and 90° (the right angle). The shortest leg (length 1) is opposite the 30° angle; the longer leg (length √3) is opposite the 60° angle; the hypotenuse (length 2) is opposite the right angle. Side ratio: 1 : √3 : 2. [1 mark]
Part (b) — Exact values. Using the 30-60-90 triangle from (a):
For the 30° angle — opposite = 1, adjacent = √3, hypotenuse = 2: [1 mark]
sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3 = √3/3 (rationalised). [1 mark]
For the 60° angle — opposite = √3, adjacent = 1, hypotenuse = 2:
sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3. [1 mark]
(Note the complementarity: sin 60° = cos 30° and sin 30° = cos 60°, reflecting the general identity sin θ = cos(90° − θ).)
Part (c) — Evaluating the expression. Substitute exact values from (b):
sin 60° + cos 30° = √3/2 + √3/2 = 2 × √3/2 = √3. [1 mark]
(√3) × tan 30° = √3 × √3/3 = (√3)² / 3 = 3/3 = 1. [1 mark] ▮
Total: 8/8.
Band descriptors for marker.
Band 3: Quotes the ratio 1:√3:2 in (a) without Pythagoras derivation; gives correct exact values in (b) but without showing the connection to the triangle; (c) computed but final answer un-simplified (e.g. leaves it as 3/3). ≈ 3-4 marks.
Band 4: Pythagoras applied correctly in (a); (b) and (c) correct; minor presentation slips (e.g. un-rationalised 1/√3 retained). ≈ 5-6 marks.
Band 5: Full derivation in (a); explicit identification of opp/adj/hyp for each angle in (b); (c) correct. Missing or weak complementarity comment. ≈ 7 marks.
Band 6: Diagram included; symmetry justification clear; all rationalisations explicit; complementarity sin 60° = cos 30° noted; (c) reaches the unexpectedly clean "= 1" with clear arithmetic. 8/8. Top scripts also note that the expression in (c) is in fact sin 60° tan 30° + cos 30° tan 30° = sin 60° tan 30° + sin 30°, exposing a hidden structure.