What does Area 2 Relationships cover in SQA National 5 Mathematics?

Area 2 is the algebra, geometry and trigonometry heart of the course. It covers the equation of a straight line and functional notation, solving equations and inequations, simultaneous equations, changing the subject of a formula, quadratic functions and equations with the discriminant, Pythagoras theorem, the angle properties of shapes and similarity, and trigonometry including the sine, cosine and tangent graphs, solving trig equations, and the sine and cosine rules with the area of a triangle. It is the largest of the three areas.

When do you use the sine rule and when the cosine rule in National 5 Maths?

Use the sine rule when you have a matching side and opposite angle pair plus one further measurement, to find a side or an angle. Use the cosine rule when you have two sides and the included angle to find the third side, or all three sides to find an angle. Neither is needed for a right-angled triangle, where Pythagoras and basic trigonometry (sine, cosine, tangent) are quicker.

What does the discriminant tell you in National 5 Maths?

The discriminant is the expression b squared minus 4ac from the quadratic formula. If it is positive there are two distinct real roots and the parabola crosses the x-axis twice; if it is zero there is one repeated real root and the parabola touches the x-axis; if it is negative there are no real roots and the parabola misses the x-axis. It tells you the number and nature of the roots without solving the equation.

How do you solve a trigonometric equation in National 5 Maths?

Isolate the trig ratio so the equation reads sine, cosine or tangent of the angle equals a number, find the first angle (often an exact value such as 30, 45 or 60 degrees), then use the symmetry of the graph to find every other solution within the stated range. Sine is positive in the first and second quadrants, cosine in the first and fourth, and tangent in the first and third, which gives the second solution in a zero to 360 degree range.

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SQA National 5 Mathematics Area 2 Relationships: the straight line, functions, equations, quadratics, geometry and trigonometry

A deep-dive SQA National 5 Mathematics guide to Area 2 Relationships. Covers the straight line and functional notation, equations and inequations, simultaneous equations, changing the subject, quadratic functions and the discriminant, Pythagoras, properties of shapes and similarity, and trigonometry including graphs, equations and the sine and cosine rules.

Generated by Claude Opus 4.818 min readNational 5Updated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What Area 2 actually demands
The straight line and functional notation
Equations, inequations and simultaneous equations
Quadratics and the discriminant
Geometry: Pythagoras, shapes and similarity
Trigonometry
How Area 2 is examined
Check your knowledge

What Area 2 actually demands

Relationships is the largest area of National 5 Mathematics and pulls together algebra, geometry and trigonometry. The examiners test fluent equation solving, clear geometric reasoning, and correct choice of method in trigonometry. This guide walks through all the topics of the area, then sets out the patterns the SQA repeats. Each topic has a matching dot-point page with practice questions; this overview ties them together.

The straight line and functional notation

The area opens with the straight line $y = mx + c$ , where $m$ is the gradient and $c$ the y-intercept, and the point-gradient form $y - b = m(x - a)$ for building an equation from a gradient and a point. Functional notation $f(x)$ names a rule; you evaluate it by substitution and solve $f(x) = k$ for an input.

Equations, inequations and simultaneous equations

Equations and inequations are solved by balancing, with the special rule that an inequality reverses when you divide by a negative. Simultaneous equations solve two linear equations in two unknowns by elimination or substitution, geometrically the point of intersection. Changing the subject rearranges a formula to isolate a chosen variable, undoing operations in reverse, including squares and roots.

Quadratics and the discriminant

Quadratic functions and equations graph as parabolas; solve them by factorising or the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . The discriminant $b^2 - 4ac$ tells you the number and nature of the roots: positive gives two roots, zero gives one repeated root, negative gives none.

Geometry: Pythagoras, shapes and similarity

Pythagoras $c^2 = a^2 + b^2$ finds a missing side of a right-angled triangle and tests for a right angle. Properties of shapes and angles covers angle sums, parallel-line angles and circle facts such as the semicircle right angle. Similarity uses the linear scale factor for lengths, its square for areas and its cube for volumes.

Trigonometry

Trigonometric graphs and equations describe the sine, cosine and tangent waves, their amplitude and period, and how to solve simple trig equations using exact values and symmetry. The sine and cosine rule solve any non-right-angled triangle, with the area formula $\tfrac{1}{2}ab\sin C$ for two sides and the included angle.

How Area 2 is examined

A typical SQA profile for Relationships:

Algebraic fluency. Solving equations, rearranging formulae and handling quadratics appear throughout.
Geometric reasoning. Angle chases and similarity reward stating the property used at each step.
Method choice in trigonometry. Picking the sine rule, cosine rule or basic trigonometry correctly is a recurring skill.

Check your knowledge

A mix of recall and method questions covering Area 2. Attempt them, then check against the solutions.

Find the equation of the line through $(1, 2)$ with gradient $4$ . (2 marks)
Solve $3x - 5 = x + 7$ . (2 marks)
Solve $x^2 - 5x + 6 = 0$ . (2 marks)
State the nature of the roots of $x^2 + 2x + 5 = 0$ . (2 marks)
Find the area of a triangle with sides $4$ cm and $6$ cm and included angle $30^\circ$ . (2 marks)

Solutions to check-your-knowledge questions

Step 1: Find the equation of a straight line from a point and gradient

Use the point-gradient form $y - b = m(x - a)$ with the given point $(1, 2)$ and gradient $4$ , then rearrange to slope-intercept form:

y - 2 = 4(x - 1) \implies y = 4x - 2.

Step 2: Solve a linear equation by collecting like terms

Gather all variable terms on one side and all constant terms on the other, then divide by the coefficient of $x$ :

3x - x = 7 + 5 \implies 2x = 12 \implies x = 6.

Step 3: Solve a quadratic equation by factorising

Factorise the left side by finding two numbers that multiply to $6$ and add to $-5$ , then set each factor equal to zero:

(x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3.

Step 4: Determine the nature of roots using the discriminant

Compute $b^2 - 4ac$ with $a = 1$ , $b = 2$ , $c = 5$ . A negative discriminant means the parabola does not cross the $x$ -axis, so there are no real roots:

b^2 - 4ac = 4 - 20 = -16 < 0 \implies \text{no real roots.}

Step 5: Find the area of a triangle using two sides and the included angle

Apply the formula $\tfrac{1}{2}ab\sin C$ with $a = 4$ cm, $b = 6$ cm, and included angle $C = 30^\circ$ , using the exact value $\sin 30^\circ = 0.5$ :

\text{Area} = \tfrac{1}{2} \times 4 \times 6 \times \sin 30^\circ = 12 \times 0.5 = 6 \text{ cm}^2.

Sources & how we know this

SQA National 5 Mathematics Course Specification — SQA (2018)