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SEC 2027 · K310 SEC 3

Sec 3 E Math
Visual Notes

Pictures first. Formulas boxed. Every exam keyword highlighted.
HOW TO STUDY WITH THIS
Read the 💡 box
the whole topic in 1 line
Study the diagram
cover it, redraw it
Learn the yellow
formulas & keywords
Redo the worked example
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THE CHAPTERS
⛰️
CH 1
Quadratic Equations
Factorise, formula, complete the square.
↔️
CH 2
Simultaneous Inequalities
Both must be true → overlap.
⚡
CH 3
Indices & Standard Form
Same base: × add, ÷ subtract.
📏
CH 4
Coordinate Geometry
Gradient, length, y = mx + c.
〰️
CH 5
Graphs of Functions
Know the 6 shapes.
📐
CH 6
Sine & Cosine Rules
Any triangle, not just right-angled.
🧭
CH 7
Bearings & 3D
From North, clockwise, 3 digits.
🍕
CH 8
Radians, Arcs & Sectors
s = rθ, A = ½r²θ.
⭕
CH 9
Circle Properties
8 theorems, always give reasons.
🔺
CH 10
Congruence & Similarity Tests
SSS SAS ASA RHS · AA.
🔍
CH 11
Similar Areas & Volumes
Area k², volume k³.
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E Math · Sec 3 · SEC K310
⛰️
1

Quadratic Equations & Functions

N6 · N7 · formula, completing the square, graphs
BIG IDEAMake it ax² + bx + c = 0 first. If it won't factorise, use the formula or complete the square.
📜 Quadratic formula
x = [−b ± √(b² − 4ac)] / 2a
2x² − 3x − 4 = 0: a = 2, b = −3, c = −4
x = [3 ± √41] / 4 = 2.35 or −0.851
🟦 Completing the square
x² + bx = (x + b/2)² − (b/2)²
x² − 6x + 2 = 0
(x − 3)² − 9 + 2 = 0 → (x − 3)² = 7
x = 3 ± √7 = 5.65 or 0.354

3 forms of the same parabola

xyturning pt (1, −4)
y = (x − 1)² − 4 = (x + 1)(x − 3)
FormTells you
y = (x − h)² + kturning point (h, k), symmetry x = h
y = (x − p)(x − q)x-intercepts p, q
y = ax² + bx + cy-intercept c

Fractional equations → quadratic

Example
3/x + 2/(x − 1) = 4
× x(x − 1): 3(x − 1) + 2x = 4x(x − 1) → 4x² − 9x + 3 = 0 → x = 1.84 or 0.407
⚠ TRAPFormula: put negatives in brackets, e.g. b = −3 → −b = +3, b² = 9. Final answers to 3 s.f..
🧠 HOOKGraphical solution: roots of ax² + bx + c = 0 are where y = ax² + bx + c cuts the x-axis. To solve f(x) = k, draw y = k and read the x-values.
E Math · Sec 3
↔️
2

Simultaneous Linear Inequalities

N7
BIG IDEASolve each inequality separately, then take the overlap — both must be true.
Example
3x − 1 < x + 7 ≤ 4x + 4
Split: 3x − 1 < x + 7 → x < 4    and    x + 7 ≤ 4x + 4 → x ≥ 1
→ 1 ≤ x < 4
-3-2-1012345
1 ≤ x < 4 · integers: 1, 2, 3
Integer / prime answers
"integer values" → list them
"prime values" of 1 ≤ x < 9 → 2, 3, 5, 7
⚠ TRAPSplit a "double" inequality into two — never solve both parts in one go if x appears in 3 places. ÷ negative → flip.
🧠 HOOKNo overlap → "no solution". Check the end points: ≤ includes, < does not.
E Math · Sec 3
⚡
3

Indices & Standard Form

N1 · laws of indices, standard form, compound interest
BIG IDEASame base: × → add powers, ÷ → subtract powers, power of power → multiply.
LawRuleExample
1aᵐ × aⁿ = aᵐ⁺ⁿ2³ × 2⁴ = 2⁷
2aᵐ ÷ aⁿ = aᵐ⁻ⁿ5⁶ ÷ 5² = 5⁴
3(aᵐ)ⁿ = aᵐⁿ(x²)³ = x⁶
4aⁿ × bⁿ = (ab)ⁿ2³ × 5³ = 10³
5aⁿ ÷ bⁿ = (a/b)ⁿ(x/y)² = x²/y²
zeroa⁰ = 17⁰ = 1
negativea⁻ⁿ = 1/aⁿ2⁻³ = ⅛
fractiona^(1/n) = ⁿ√a8^(1/3) = 2
fractiona^(m/n) = (ⁿ√a)ᵐ27^(2/3) = 3² = 9
Solve with indices
make the bases equal, then equate powers
2ˣ = 32 → 2ˣ = 2⁵ → x = 5
9ˣ = 27 → 3²ˣ = 3³ → x = 1.5
🔬 Standard form
A × 10ⁿ, 1 ≤ A < 10
345 000 = 3.45 × 10⁵
0.00072 = 7.2 × 10⁻⁴
(3 × 10⁴)(5 × 10³) = 15 × 10⁷ = 1.5 × 10⁸
💰 Compound interest (given)
A = P(1 + r/100)ⁿ
$5000 at 2% p.a. for 3 yrs → A = 5000(1.02)³ = $5306.04
interest = A − P = $306.04
⚠ TRAPCompounded monthly: rate ÷ 12, n = years × 12. Standard form: A must be between 1 and 10 — 15 × 10⁷ is not standard form.
E Math · Sec 3
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4

Coordinate Geometry

G6 · gradient, length, equation of a line
BIG IDEAFor points A(x₁, y₁), B(x₂, y₂): gradient, length and equation all come from the differences.
xyA(1, 1)B(4, 7)Δx = 3Δy = 6
y = 2x − 1
3 formulas
m = (y₂ − y₁) / (x₂ − x₁)
= 6 ÷ 3 = 2
AB = √[(x₂ − x₁)² + (y₂ − y₁)²]
= √(9 + 36) = 6.71
y − y₁ = m(x − x₁)
y − 1 = 2(x − 1) → y = 2x − 1
Parallel lines
same gradient
line ∥ to y = 3x + 2 through (0, −5): y = 3x − 5
Where lines meet
x-axis: put y = 0 · y-axis: put x = 0
two lines: solve simultaneously
Area of a shape
split into triangles / trapeziums, or use base × height ÷ 2 with a horizontal or vertical side
⚠ TRAPKeep the same order on top and bottom of the gradient formula. Length is always positive.
E Math · Sec 3
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5

Graphs of Functions

N6 · power, reciprocal, exponential graphs, tangents
BIG IDEARecognise each graph by its shape. Then use the graph to read values and solve equations.
y = ax³ (a>0)
y = ax³ (a<0)
y = a/x
y = a/x²
y = kaˣ (a>1)
y = ka⁻ˣ
y = x² − 1
y = x + 1/x (sum)
GraphKey feature
y = a/xtwo branches, never touches axes (asymptotes x = 0, y = 0)
y = a/x²both branches above x-axis when a > 0
y = kaˣcuts y-axis at k, never touches x-axis
y = ax³S-shape through origin

Gradient of a curve — draw a tangent

xy(2, 2)
tangent touches at one point
Steps
① draw a straight line that just touches the curve
② pick 2 far-apart points on the tangent
③ gradient = rise ÷ run
here: 4 ÷ 2 = 2
🧠 HOOKSolve f(x) = 0 graphically: read where the curve cuts y = 0. Solve f(x) = g(x): draw both, read the x-values where they meet.
E Math · Sec 3
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6

Sine & Cosine Rules

G4 · obtuse angles, area ½ab sin C
BIG IDEAFor any triangle: side a is opposite angle A. Choose sine rule or cosine rule by what you are given.
BCA abc
small letter = side opposite capital angle
Formulas (given in exam)
a/sin A = b/sin B = c/sin C
a² = b² + c² − 2bc cos A
Area = ½ab sin C
You know…UseFind
2 angles + 1 sidesine ruleanother side
2 sides + non-included anglesine rulean angle
2 sides + included anglecosine rule3rd side
all 3 sidescosine rulecos A = (b² + c² − a²)/2bc
Cosine example
b = 7, c = 5, A = 40°
a² = 49 + 25 − 70 cos 40° = 20.38
a = 4.51 · area = ½(7)(5) sin 40° = 11.2
Sine example
A = 50°, B = 60°, a = 8
b = 8 sin 60° ÷ sin 50° = 9.04

Obtuse angles (90° – 180°)

Rules
sin(180° − θ) = sin θ
cos(180° − θ) = −cos θ
cos is negative for obtuse angles → cosine rule gives obtuse angles automatically
⚠ TRAPsin⁻¹ only gives the acute angle. If the angle could be obtuse, also check 180° − answer.
E Math · Sec 3
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7

Bearings, Elevation & 3D

G4 · applications of trigonometry
BIG IDEABearing = angle measured from North, clockwise, written with 3 digits (e.g. 065°).
N AB051° N231°
bearing of B from A = 051° · A from B = 051° + 180° = 231°
Rules
draw a North line at every point
back bearing = bearing ± 180°
North lines are parallel → use alternate / co-interior angles

Elevation & depression

elevation depression
both measured from the horizontal — they are equal (alt. ∠s)
🧊 3D problems
① find the right-angled triangle that contains the angle
② you often need a line on the base first (Pythagoras)
③ angle between a line and a plane = angle with its projection on the plane
⚠ TRAPAngle of depression is measured down from the horizontal, never from the vertical. Always draw the North arrow before measuring.
E Math · Sec 3
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8

Radians, Arcs & Sectors

G5 · arc length, sector & segment area
BIG IDEAπ rad = 180°. With θ in radians: arc s = rθ and sector A = ½r²θ.
rθschord
sector (green) · arc (red) · segment = between chord and arc
Formulas
degrees → rad: × π/180
rad → degrees: × 180/π
arc length s = rθ
sector area = ½r²θ
segment = ½r²(θ − sin θ)
Example
r = 6 cm, θ = 1.2 rad
s = 6(1.2) = 7.2 cm
sector = ½(36)(1.2) = 21.6 cm²
segment = 18(1.2 − sin 1.2) = 4.82 cm²
Perimeter of sector
= 2r + s = 12 + 7.2 = 19.2 cm
⚠ TRAPCalculator in RAD mode for sin 1.2. Using degrees? then use (θ/360) × 2πr and (θ/360) × πr².
🧠 HOOKCommon: π/6 = 30°, π/4 = 45°, π/3 = 60°, π/2 = 90°, 2π = 360°.
E Math · Sec 3
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9

Properties of Circles

G3 · symmetry & angle properties
BIG IDEA8 circle facts. In every proof, write the reason next to each step.
⊥ from centre bisects chord
equal chords are equidistant from centre
tangent ⊥ radius
∠ in semicircle = 90°
2xx
∠ at centre = 2 × ∠ at circumference
∠s in same segment are equal
cyclic quad: opposite ∠s = 180°
==
tangents from a point are equal
✍ REASONSUse short reasons: "∠ in semicircle", "∠ at centre = 2∠ at circumference", "∠s in same segment", "∠s in opp. segment" (cyclic quad), "tan ⊥ rad", "tangents from ext. pt".
⚠ TRAPLook for hidden isosceles triangles: two radii from the centre are equal → base angles equal.
E Math · Sec 3
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10

Congruence & Similarity Tests

G2 · tests, proofs, bisectors, scale drawings
BIG IDEAProve congruence/similarity by listing the matching parts with reasons, then state the test.
≡ Congruence tests
SSSSASASA / AASRHS
RHS = right angle, hypotenuse, side
AAA and SSA do NOT prove congruence
~ Similarity tests
AASSS (all in ratio)SAS (ratio + included ∠)
AA is the easiest: 2 equal angles
Proof layout
In △ABC and △DEF:
AB = DE (given) · ∠B = ∠E (alt. ∠s, AB ∥ DE) · BC = EF (given)
∴ △ABC ≡ △DEF (SAS)

Bisectors & scale drawings

AB
⊥ bisector: points equidistant from A and B
∠ bisector: points equidistant from both lines
🧠 HOOKScale drawing: write the scale, use a sharp pencil, ruler and protractor, keep construction arcs. Measure to the nearest mm / degree.
E Math · Sec 3
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11

Areas & Volumes of Similar Figures

G2
BIG IDEAIf lengths are in ratio k, areas are in ratio k² and volumes (and masses) in ratio k³.
1 2 → area × 4 volume × 8
double the length → area ×4, volume ×8
Ratio of…Rule
lengthsl₁ : l₂ = k
areasA₁/A₂ = (l₁/l₂)²
volumes / massesV₁/V₂ = (l₁/l₂)³
Example
Similar cones, heights 3 : 5. Small volume = 54 cm³
V₂ = 54 × (5/3)³ = 250 cm³
area ratio = 9 : 25
Going backwards
areas 16 : 25 → lengths √ → 4 : 5
volumes 8 : 27 → lengths ∛ → 2 : 3 → areas 4 : 9
⚠ TRAPNever use the area ratio for volumes. Always go back to the length ratio first, then square or cube.
🧠 HOOKTriangles with the same height: area ratio = base ratio (useful in similarity questions).
E Math · Sec 3