IGCSE 0580

Board · Cambridge IGCSE Mathematics (0580) Tiers · Core & Extended Chapters · 8 Explorers · 3 interactive

Number

📌 0580 Topic 1 — Number C E · Types of number · Fractions, decimals, percentages · Indices & standard form · Surds E · Ratio & proportion · Financial maths · Limits of accuracy E

🗺 Cambridge IGCSE 0580 — Syllabus Map

C Core topics are examinable on Papers 1 & 3 (Core tier). E Extended topics appear on Papers 2 & 4 only.

ClipSAT Chapter	0580 Topics covered	Tier
1 Number	Types of number, fractions, %, ratio & proportion, standard form, surds, indices, financial maths, limits of accuracy	C E
2 Algebra & Graphs	Expressions, equations, sequences, inequalities, simultaneous equations, graph types	C E
3 Coordinate Geometry	Gradient, midpoint, distance, equation of a line, perpendicular lines	C E
4 Mensuration	Perimeter, area, arc, sector, surface area, volume of 3D shapes	C E
5 Geometry	Angles, polygons, circle theorems, similarity & congruence, construction & loci	C E
6 Trigonometry	SOH-CAH-TOA, sine/cosine rules, area of triangle, bearings, 3D trig	C E
7 Transformations & Vectors	Translation, reflection, rotation, enlargement, column vectors, vector paths, magnitude	C E
8 Probability & Statistics	Probability, tree diagrams, conditional probability, averages, spread, frequency tables, histograms, cumulative frequency, box plots	C E
9 Functions	Function notation, domain, composite, inverse	E
10 Sets & Venn Diagrams	Set notation, union, intersection, complement, two- and three-set Venn diagrams	C E

The numerical backbone of the whole syllabus: working confidently with fractions, percentages, ratio, standard form, indices and rounding. Most marks here are earned by careful, methodical arithmetic rather than clever tricks.

● Definition 1.1 — Standard form

A number is in standard form when written as $a\times 10^{n}$ with $1\le a<10$ and $n$ an integer. Large numbers take a positive $n$ ; small numbers (less than 1) take a negative $n$ .

Percentage change

\dfrac{\text{change}}{\text{original}}\times 100\%

Index law 1 — Product

a^{m}\,a^{n}=a^{m+n}

Index law 2 — Quotient

\dfrac{a^{m}}{a^{n}}=a^{m-n}

Index law 3 — Power

(a^{m})^{n}=a^{mn}

Reverse %

\text{original}=\dfrac{\text{new value}}{1\pm r}

Bounds

value rounded to a unit lies within

\pm\tfrac12

unit

Worked example 1.A

Write $0.000452$ in standard form.

Move the decimal point four places right to land between 4 and 5, so the exponent is $-4$ :

0.000452 = 4.52\times 10^{-4}.

Worked example 1.B

A coat costs $60 after a 20% discount. Find its original price.

The sale price is 80% of the original, so divide rather than add back 20%:

\text{original}=\frac{60}{0.8}=\$75.

● Surds E

A surd is an irrational root left in exact form (e.g. $\sqrt{3}$ ). Key rules: $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ ; $\dfrac{1}{\sqrt{a}}=\dfrac{\sqrt{a}}{a}$ (rationalising a denominator). To rationalise $\dfrac{p}{a+\sqrt{b}}$ , multiply numerator and denominator by the conjugate $a-\sqrt{b}$ .

Surd product

\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}

Rationalise

\dfrac{1}{\sqrt{a}}=\dfrac{\sqrt{a}}{a}

Compound interest

A=P\!\left(1+\dfrac{r}{100}\right)^{\!n}

Limits of accuracy

lower =

x-\tfrac{1}{2}\delta

, upper =

x+\tfrac{1}{2}\delta

Worked example 1.C — Surds

Simplify $\sqrt{48}-\sqrt{27}$ .

\sqrt{48}=4\sqrt{3},\quad\sqrt{27}=3\sqrt{3}.

\sqrt{48}-\sqrt{27}=4\sqrt{3}-3\sqrt{3}=\sqrt{3}.

Worked example 1.D — Limits of accuracy

A length is measured as $12.4$ cm, rounded to 1 d.p. State the upper and lower bounds.

Half the smallest unit = 0.05.
Lower bound: $12.35$ cm. Upper bound: $12.45$ cm.

£800 is invested at 3.5% compound interest per year. Find the amount after 4 years, to the nearest penny.

A=800\left(1+\frac{3.5}{100}\right)^4=800\times1.035^4\approx800\times1.1475=£918.00

Number line representation of the inequality

-1 \le x < 4

, with solid point for inclusive bound and open circle for strict bound.

⚠️ Cambridge 0580 Traps — Significant Figures & Bounds

Significant Figures vs. Decimal Places: Unless specified otherwise, Cambridge IGCSE demands answers to 3 significant figures (or 1 d.p. for angles). Leading zeros are NOT significant: $0.00350$ has 3 s.f., while $350$ is ambiguous unless written in standard form $3.50 \times 10^2$ .

Limits of Accuracy: For division $\frac{A}{B}$ , the maximum possible value occurs when the numerator is at its upper bound and the denominator is at its lower bound: $\text{Max} = \frac{A_{\text{upper}}}{B_{\text{lower}}}$ . Never round intermediate values during multi-step bound calculations.

A car travels a distance of

d = 180\text{ km}

correct to the nearest 10 km in a time of

t = 2.4\text{ hours}

correct to 1 decimal place. Calculate the upper bound for the average speed of the car in km/h.

Find the bounds for distance and time:

d_{\text{upper}} = 180 + 5 = 185\text{ km}

t_{\text{lower}} = 2.4 - 0.05 = 2.35\text{ hours}

To maximize speed $s = \frac{d}{t}$ , divide the upper distance by the lower time:

s_{\text{upper}} = \frac{185}{2.35} \approx 78.723\dots \approx 78.7\text{ km/h (3 s.f.)}

📝 Chapter Quiz

Questions Level

Algebra & Graphs

📌 0580 Topic 2 — Algebra & Graphs C E · Expressions & equations · Sequences · Inequalities · Simultaneous equations · Graph recognition & sketching · Kinematics graphs E

Expanding, factorising, solving and rearranging — plus reading and sketching graphs. This is the strand that feeds directly into AS and A-Level, so the habits you build here matter well beyond the exam.

● Definition 2.1 — Factorising a quadratic

To factorise $x^{2}+bx+c$ , find two numbers that multiply to $c$ and add to $b$ . They become the constants in $(x+p)(x+q)$ .

Difference of squares

a^{2}-b^{2}=(a-b)(a+b)

Quadratic formula

x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}

Arithmetic nth term

a+(n-1)d

Rearranging

apply inverse operations to both sides

Worked example 2.A

Solve the simultaneous equations $x+y=7$ and $2x-y=2$ .

Adding eliminates $y$ : $3x=9$ , so $x=3$ . Substitute back into the first equation:

x=3,\qquad y=4.

Worked example 2.B

Find the $n$ th term of $5,\,8,\,11,\,14,\ldots$

The common difference is $3$ and the term before the first would be $2$ :

u_n=3n+2.

● Sequences & nth Term

An arithmetic sequence increases/decreases by a fixed common difference $d$ . The $n$ th term is $u_n = a+(n-1)d$ where $a$ is the first term. A geometric sequence multiplies by a common ratio $r$ : $u_n=ar^{n-1}$ .

Arithmetic

u_n

a+(n-1)d

Geometric

u_n

ar^{n-1}

Linear inequality

Flip sign when multiplying/dividing by negative

Graph types

linear, quadratic (∪∩), cubic (S), reciprocal (1/x)

Worked example 2.C — nth term

Find the $n$ th term of: 5, 9, 13, 17, …

$d=4$ , first term $a=5$ . $u_n=5+(n-1)\times4=4n+1$ .

● Linear Inequalities & Graph Recognition E

Solve like an equation but reverse the inequality sign when multiplying or dividing by a negative number. Graph types you must recognise: linear $y=mx+c$ (straight line), quadratic $y=ax^2+bx+c$ (parabola), cubic $y=ax^3+\ldots$ (S-shape), reciprocal $y=\dfrac{k}{x}$ (two branches), exponential $y=a^x$ (crosses $y$ -axis at 1).

Worked example 2.D — Linear inequality

Solve $3-2x<11$ and represent on a number line.

-2x<8\;\Rightarrow\;x>-4.

Open circle at $-4$ , arrow pointing right.

The

n

th term of a sequence is

3n-7

. (a) Find the 10th term. (b) Which term has value 23?

(a)

3(10)-7=23

. (b)

3n-7=23\Rightarrow3n=30\Rightarrow n=10

. Term 10 has value 23.

Quadratic graph

y = x^2 - 4x + 3

intersecting the x-axis at roots

x = 1

and

x = 3

, with axis of symmetry

x = 2

and vertex minimum at

(2, -1)

⚠️ Cambridge 0580 Traps — Algebraic Operations & Inequalities

Sign Flips in Inequalities: When dividing or multiplying by a negative number, the inequality sign MUST reverse: $-3x \le 12 \Rightarrow x \ge -4$ . Forgetting this flip is the #1 lost mark in Cambridge algebra.

Expanding Binomial Squares: Remember that $(a - b)^2 = a^2 - 2ab + b^2$ , NOT $a^2 - b^2$ . Always write out $(2x - 3)(2x - 3)$ fully to avoid omitting the middle term $-12x$ .

Quadratic Formula Denominator: The division line extends across the ENTIRE numerator: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Do not calculate $-b \pm \frac{\sqrt{\dots}}{2a}$ .

Solve the quadratic equation

2x^2 + 5x - 4 = 0

. Give your solutions correct to 2 decimal places.

Identify coefficients: $a = 2$ , $b = 5$ , $c = -4$ .

x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-4)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 32}}{4} = \frac{-5 \pm \sqrt{57}}{4}

x_1 = \frac{-5 + 7.5498}{4} = \frac{2.5498}{4} \approx 0.64

x_2 = \frac{-5 - 7.5498}{4} = \frac{-12.5498}{4} \approx -3.14

Solutions: $x = 0.64$ or $x = -3.14$ (both to 2 d.p.).

📝 Chapter Quiz

Questions Level

Coordinate Geometry

📌 0580 Topic 3 — Coordinate Geometry C E · Gradient · Midpoint & distance · Equation of a line · Parallel & perpendicular lines

Everything about straight lines on the $xy$ -plane: gradient, length, midpoint and the equation $y=mx+c$ . Parallel lines share a gradient; perpendicular gradients multiply to $-1$ .

▲ Reference — Coordinate Geometry

For points $(x_1,y_1)$ and $(x_2,y_2)$ : gradient $m=\dfrac{y_2-y_1}{x_2-x_1}$ ; midpoint $=\!\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$ ; distance $=\!\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . Two lines are parallel if they have equal gradients. Two lines are perpendicular if the product of their gradients is $-1$ : $m_1\times m_2=-1$ .

Perpendicular gradient

m_\perp=-\dfrac{1}{m}

Parallel lines

m_1=m_2

General line

ax+by=c

Intercepts

Set

y=0

x=0

Worked example 3.B — Perpendicular line

Line $\ell$ has equation $y=3x-2$ . Find the equation of the line perpendicular to $\ell$ passing through $(6,1)$ .

Perpendicular gradient: $m_\perp=-\tfrac{1}{3}$ .

y-1=-\tfrac{1}{3}(x-6)\;\Rightarrow\;y=-\tfrac{1}{3}x+3.

Show that the points

A(1,3)

B(4,7)

and

C(7,3)

form an isosceles triangle.

AB=\sqrt{9+16}=5

BC=\sqrt{9+16}=5

AC=\sqrt{36+0}=6

. Since

AB=BC

, triangle is isosceles.

Gradient

m=\dfrac{y_2-y_1}{x_2-x_1}

Midpoint

\Big(\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2}\Big)

Distance

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Straight line

y=mx+c

Worked example 3.A

Find the equation of the line through $(1,2)$ and $(4,11)$ .

Gradient $m=\dfrac{11-2}{4-1}=3$ . Use $(1,2)$ : $2=3(1)+c \Rightarrow c=-1$ :

y=3x-1.

Linear graph

y = 2x - 1

with gradient

m = 2

and y-intercept

(0, -1)

⚠️ Cambridge 0580 Traps — Coordinate Geometry

Gradient Order: Gradient is $m = \frac{y_2 - y_1}{x_2 - x_1}$ . Always keep the coordinate order consistent; subtracting $y_2 - y_1$ but $x_1 - x_2$ causes a fatal sign error.

Perpendicular Gradient: For a perpendicular line, take both the negative and the reciprocal. If $m = -\frac{3}{4}$ , then $m_\perp = +\frac{4}{3}$ . Students frequently forget the minus sign or forget to invert.

Rearranging to Standard Form: Before reading off the gradient from $3x + 2y = 8$ , always rearrange to $y = mx + c$ : $2y = -3x + 8 \Rightarrow y = -\frac{3}{2}x + 4$ , so $m = -\frac{3}{2}$ , NOT 3.

Find the equation of the perpendicular bisector of the line segment joining

P(2, -1)

and

Q(8, 7)

. Give your answer in the form

y = mx + c

Step 1: Find the midpoint $M$ of $PQ$ :

M = \left(\frac{2 + 8}{2}, \frac{-1 + 7}{2}\right) = (5, 3)

Step 2: Find the gradient of line $PQ$ :

m_{PQ} = \frac{7 - (-1)}{8 - 2} = \frac{8}{6} = \frac{4}{3}

Step 3: Determine the perpendicular gradient:

m_\perp = -\frac{3}{4}

Step 4: Form the equation passing through $M(5, 3)$ :

y - 3 = -\frac{3}{4}(x - 5) \implies y - 3 = -\frac{3}{4}x + \frac{15}{4}

y = -\frac{3}{4}x + \frac{27}{4} \quad \text{or} \quad y = -0.75x + 6.75

📝 Chapter Quiz

Questions Level

Mensuration

📌 0580 Topic 4 — Mensuration C E · Perimeter & area · Arc length & sector area · Volume & surface area of prisms, cylinders, cones, spheres

Perimeter, area, surface area and volume — from circles and trapezia up to cylinders, cones and spheres. Read each question carefully for the value of $\pi$ it expects (often $3.14$ or $\tfrac{22}{7}$ ).

▲ Reference — The circle

Circumference $C=2\pi r=\pi d$ ; area $A=\pi r^{2}$ . An arc and sector are simply fractions of these, scaled by $\dfrac{\theta}{360}$ .

Circle

C=2\pi r,\quad A=\pi r^{2}

Cylinder

V=\pi r^{2}h

Sphere

V=\tfrac{4}{3}\pi r^{3}

Cone

V=\tfrac{1}{3}\pi r^{2}h

Prism

V=A_{\text{cross}}\times \ell

Trapezium

A=\tfrac12(a+b)h

Worked example 4.A

Find the volume of a cylinder with $r=3$ cm and $h=10$ cm. Take $\pi=3.14$ .

V=\pi r^{2}h=3.14\times 9\times 10=282.6\ \text{cm}^{3}.

Worked example 4.B

Find the volume of a sphere of radius $3$ cm. Take $\pi=3.14$ .

V=\tfrac{4}{3}\pi r^{3}=\tfrac{4}{3}\times 3.14\times 27=113.04\ \text{cm}^{3}.

Right circular cylinder with radius

r = 3\text{ cm}

and height

h = 10\text{ cm}

⚠️ Cambridge 0580 Traps — Mensuration & Units

Area & Volume Unit Conversions: Converting square and cubic units is a major Cambridge trap. $1\text{ m}^2 = (100\text{ cm})^2 = 10,000\text{ cm}^2$ , NOT 100! Similarly, $1\text{ m}^3 = 1,000,000\text{ cm}^3$ and $1\text{ litre} = 1,000\text{ cm}^3$ .

Total Surface Area: For open-topped or hollow solids, omit missing faces. A closed cylinder has area $2\pi rh + 2\pi r^2$ ; a solid hemisphere has total surface area $2\pi r^2 + \pi r^2 = 3\pi r^2$ (including the flat circular base).

Slant Height vs. Vertical Height: In cones and pyramids, $h$ is the perpendicular vertical height and $l$ is the slant height, linked by Pythagoras: $l = \sqrt{r^2 + h^2}$ . The curved surface area of a cone uses $l$ ( $\pi r l$ ), while volume uses $h$ ( $\frac{1}{3}\pi r^2 h$ ).

Worked example 4.C — Sector Area & Arc Length

A sector of a circle has radius $r = 6\text{ cm}$ and sector angle $\theta = 120^\circ$ . Calculate (a) the arc length and (b) the area of the sector, giving answers in terms of $\pi$ .

(a) Arc length: $L = \frac{120}{360} \times 2\pi (6) = \frac{1}{3} \times 12\pi = 4\pi\text{ cm}$ .

(b) Sector area: $A = \frac{120}{360} \times \pi (6^2) = \frac{1}{3} \times 36\pi = 12\pi\text{ cm}^2$ .

A solid toy is formed by attaching a cone of radius 4 cm and height 9 cm to the flat face of a solid hemisphere of radius 4 cm. Calculate the total volume of the toy. Give your answer correct to 3 significant figures.

Step 1: Volume of cone:

V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (4^2)(9) = 48\pi\text{ cm}^3

Step 2: Volume of hemisphere:

V_{\text{hemi}} = \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi (4^3) = \frac{128}{3}\pi\text{ cm}^3

Step 3: Total volume:

V_{\text{total}} = 48\pi + \frac{128}{3}\pi = \frac{272}{3}\pi \approx 284.838\dots \approx 285\text{ cm}^3\text{ (3 s.f.)}

📝 Chapter Quiz

Questions Level

Geometry

📌 0580 Topic 5 — Geometry C E · Angle properties · Polygons · Circle theorems E · Similarity & congruence · Construction & loci

Angle reasoning is the heart of IGCSE geometry: angles on lines and in polygons, and the family of circle theorems. Every answer should be justified with a reason, not just a number.

▲ Reference — Polygons

Interior angles of an $n$ -sided polygon sum to $(n-2)\times 180^{\circ}$ . The exterior angles always sum to $360^{\circ}$ , so a regular polygon has each exterior angle $\dfrac{360^{\circ}}{n}$ .

▲ Reference — Circle theorems

① The angle at the centre is twice the angle at the circumference on the same arc. ② The angle in a semicircle is $90^{\circ}$ . ③ Angles in the same segment are equal. ④ Opposite angles of a cyclic quadrilateral sum to $180^{\circ}$ . ⑤ A tangent meets the radius at $90^{\circ}$ .

The first theorem is the one students doubt most — so watch it directly. Point $P$ slides around the major arc while $A$ and $B$ stay fixed; the inscribed angle never changes, and it is always exactly half the central angle.

⊙ Explorer · Inscribed & central angleinteractive

Slide P around the major arc

inscribed ∠APB—

central ∠AOB—

Wherever $P$ sits on the arc, $\angle APB=\tfrac12\,\angle AOB$ . The inscribed angle holds steady.

Triangle angle sum

\alpha+\beta+\gamma=180°

Exterior angle

Exterior

=

sum of two non-adjacent interior angles

Polygon angle sum

S=(n-2)\times180°

; each interior

=\frac{S}{n}

(regular)

Parallel line angles

Alternate = equal; co-interior

= 180°

; corresponding = equal

Worked example 5.A

Find each interior angle of a regular pentagon.

Interior sum $=(5-2)\times 180^{\circ}=540^{\circ}$ ; divide equally among five angles:

\frac{540^{\circ}}{5}=108^{\circ}.

Worked example 5.B

Points $A$ and $B$ lie on a circle, centre $O$ , with $\angle AOB=80^{\circ}$ . Find the angle subtended by $AB$ at the circumference.

By the centre–circumference theorem the inscribed angle is half the central angle:

\tfrac12\times 80^{\circ}=40^{\circ}.

● Similarity & Congruence

Congruent shapes are identical in shape and size (SSS, SAS, ASA, RHS). Similar shapes have the same angles; corresponding sides are in the same ratio $k$ . Area scales as $k^2$ ; volume scales as $k^3$ .

Congruence

SSS, SAS, ASA, RHS

Similar — lengths

ratio

k

Similar — areas

ratio

k^{2}

Similar — volumes

ratio

k^{3}

Worked example 5.C — Similar areas

Two similar triangles have corresponding sides in the ratio $3:5$ . If the smaller triangle has area $18\text{ cm}^2$ , find the area of the larger.

\text{Area ratio}=\left(\tfrac{5}{3}\right)^{\!2}=\tfrac{25}{9}.\qquad A_{\text{large}}=18\times\tfrac{25}{9}=50\text{ cm}^2.

● Construction & Loci

A locus is the set of all points satisfying a given condition. Key loci: (1) equidistant from a fixed point → circle; (2) equidistant from two fixed points → perpendicular bisector; (3) equidistant from two lines → angle bisector; (4) fixed distance from a line → pair of parallel lines.

Two similar cylinders have radii 4 cm and 6 cm. If the volume of the smaller is

160\pi

cm³, find the volume of the larger.

k=6/4=1.5

. Volume ratio

=1.5^3=3.375

. Volume

=160\pi\times3.375=540\pi

cm³.

Circle with center O showing central angle

\angle AOB = 80^\circ

; the inscribed angle subtended at the circumference is half this value (

40^\circ

⚠️ Cambridge 0580 Traps — Circle Theorems & Geometric Proofs

Mandatory Geometric Reasons: Cambridge exam mark schemes award separate method and reasoning marks. You must use the exact syllabus wording: 'angle at centre is twice angle at circumference', 'angles in same segment are equal', 'opposite angles of cyclic quad sum to 180°', and 'angle in a semicircle is 90°'.

Alternate Segment Theorem: The angle between a tangent and a chord at the point of contact equals the angle in the alternate segment. Look for the triangle inside the circle resting on the tangent point.

Hidden Isosceles Triangles: Two radii drawn from the centre $O$ to points on the circumference automatically form an isosceles triangle ( $OA = OB = r$ ). Always mark base angles equal before attempting further angle calculations.

In a cyclic quadrilateral

ABCD

, angle

ABC = (2x + 15)^\circ

and angle

ADC = (3x - 10)^\circ

. Find the value of

x

and hence the size of angle

ABC

. Give a geometric reason for your equation.

Geometric reason: Opposite angles of a cyclic quadrilateral sum to $180^\circ$ .

(2x + 15) + (3x - 10) = 180

5x + 5 = 180 \implies 5x = 175 \implies x = 35

Calculate angle $ABC$ :

\angle ABC = 2(35) + 15 = 70 + 15 = 85^\circ

📝 Chapter Quiz

Questions Level

Trigonometry

📌 0580 Topic 6 — Trigonometry C E · SOH-CAH-TOA & Pythagoras · Sine & cosine rules · Area $\tfrac{1}{2}ab\sin C$ · Bearings · 3D trigonometry E

Right-angled triangles via Pythagoras and SOH-CAH-TOA, then any triangle through the sine and cosine rules. Bearings questions are trigonometry in disguise — always start with a clear diagram.

● Definition 6.1 — Right-triangle ratios

For an acute angle $\theta$ in a right triangle, $\sin\theta=\dfrac{\text{opp}}{\text{hyp}}$ , $\cos\theta=\dfrac{\text{adj}}{\text{hyp}}$ , $\tan\theta=\dfrac{\text{opp}}{\text{adj}}$ — the rule remembered as SOH-CAH-TOA.

Drag the angle and watch the three ratios respond. The hypotenuse is fixed in length; the opposite and adjacent sides — and the ratios built from them — change together.

⊙ Explorer · The trig ratiosinteractive

Angle θ = 35

sin θ—

cos θ—

tan θ—

Opposite & adjacent are drawn in indigo, the hypotenuse in amber. As $\theta\to 90^{\circ}$ , $\tan\theta$ grows without bound.

Pythagoras

a^{2}+b^{2}=c^{2}

Sine rule

\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}

Cosine rule

c^{2}=a^{2}+b^{2}-2ab\cos C

Triangle area

\tfrac12\,ab\sin C

Worked example 6.A

A right triangle has hypotenuse $10$ and an angle of $30^{\circ}$ . Find the side opposite that angle.

\text{opp}=10\sin 30^{\circ}=10\times 0.5=5.

Worked example 6.B

In a triangle, $a=7$ , $b=8$ and the included angle $C=60^{\circ}$ . Find $c$ .

Apply the cosine rule with $\cos 60^{\circ}=0.5$ :

c^{2}=49+64-2(7)(8)(0.5)=57,\qquad c=\sqrt{57}.

● Bearings

A bearing is a clockwise angle measured from North, always written as a three-digit number (e.g. 045°, 270°). Draw a North line at the starting point, then measure clockwise. The bearing from B back to A is the back-bearing = bearing ± 180°.

Worked example 6.C — Bearings with trig

A ship sails 40 km on a bearing of 130° from port P. How far east and south of P is it?

East component: $40\sin50°\approx30.6$ km. South component: $40\cos50°\approx25.7$ km.

(Angle from South = 180°−130° = 50°.)

● 3D Trigonometry E

Identify a right-angled triangle within the 3D figure: find a face diagonal using Pythagoras, then use that as one side of a second right triangle. The angle of elevation is measured upward from horizontal; the angle of depression downward from horizontal.

Back-bearing

bearing

\pm180°

3D diagonal

\sqrt{l^2+w^2+h^2}

Angle of elevation

upward from horizontal

Angle of depression

downward from horizontal

Worked example 6.D — 3D trig E

A cuboid has length 8 cm, width 6 cm, height 3 cm. Find the angle the space diagonal makes with the base.

Base diagonal $=\sqrt{8^2+6^2}=10$ cm.

\tan\theta=\frac{3}{10}\;\Rightarrow\;\theta=\arctan(0.3)\approx16.7°.

A tower PQ stands vertically. From point A on level ground, 50 m from the base Q, the angle of elevation of P is 32°. Find the height of the tower.

h=50\tan32°\approx50\times0.6249\approx31.2

Right-angled triangle ABC with sides 3, 4, 5 showing adjacent, opposite, and hypotenuse relative to angle B.

⚠️ Cambridge 0580 Traps — Trigonometry & Bearings

Calculator Angle Mode: Verify your calculator is set to D (Degrees), never R (Radians) or G (Gradians). A simple check: $\sin 30^\circ = 0.5$ . If it gives $-0.988$ , your calculator is in radians!

Three-Figure Bearings: Bearings MUST always be measured clockwise from North and written with 3 digits: $065^\circ$ , not $65^\circ$ . The back-bearing from B to A is $\text{bearing} \pm 180^\circ$ .

Ambiguous Sine Rule Case: When using $\frac{\sin A}{a} = \frac{\sin B}{b}$ to find an obtuse angle, remember $\sin(180^\circ - \theta) = \sin \theta$ . Check whether the geometry requires an acute or obtuse angle.

In triangle

XYZ

XY = 7\text{ cm}

YZ = 10\text{ cm}

, and angle

XYZ = 65^\circ

. Calculate (a) the length of

XZ

to 3 significant figures and (b) the area of triangle

XYZ

(a) Use the Cosine Rule:

XZ^2 = 7^2 + 10^2 - 2(7)(10)\cos(65^\circ) = 49 + 100 - 140(0.4226) = 149 - 59.166 = 89.834

XZ = \sqrt{89.834} \approx 9.4779\dots \approx 9.48\text{ cm (3 s.f.)}

(b) Area of triangle:

\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}(7)(10)\sin(65^\circ) = 35(0.9063) \approx 31.7\text{ cm}^2\text{ (3 s.f.)}

📝 Chapter Quiz

Questions Level

Transformations & Vectors

📌 0580 Topic 7 — Transformations & Vectors C E · Translation · Reflection · Rotation · Enlargement · Combined transformations E · Vectors & vector paths · Magnitude E

Four transformations move shapes around the plane, and column vectors describe the moves precisely. A full description must name the transformation and its defining data.

● The four transformations

Translation — give the column vector. Reflection — give the mirror line. Rotation — give the centre, angle and direction. Enlargement — give the centre and scale factor (a fraction shrinks; a negative factor flips through the centre).

📊 Cambridge IGCSE Mathematics 0580 — Topic Weighting

Number

~30% CoreExtended

Algebra & Graphs

~30% Extended

Geometry & Mensuration

~20% CoreExtended

Statistics & Probability

~20% CoreExtended

Source: Cambridge Assessment International Education — Syllabus 0580 (2025–2027)

Pick a transformation and compare the faint original triangle with its solid image. The mapping rule $(x,y)\to(\ldots)$ updates underneath.

⊙ Explorer · Transforming a triangleinteractive

Transformation

Mapping: (x, y) → (x, −y)

Column vector

\begin{pmatrix} x \\ y \end{pmatrix}

Magnitude

|\mathbf{v}|=\sqrt{x^{2}+y^{2}}

Addition

add components separately

Worked example 7.A

Reflect the point $(3,2)$ in the $x$ -axis.

Reflection in the $x$ -axis sends $(x,y)\to(x,-y)$ :

(3,2)\ \longmapsto\ (3,-2).

Worked example 7.B

Find the magnitude of the vector $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ .

|\mathbf{v}|=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5.

● Vector Magnitude & Addition E

The magnitude of $\mathbf{a}=\binom{x}{y}$ is $|\mathbf{a}|=\sqrt{x^2+y^2}$ . Vectors add tip-to-tail: $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$ . To find a vector path, combine known vectors (you may need to reverse direction, which negates the vector).

Magnitude

|\mathbf{a}|=\sqrt{x^2+y^2}

Vector path

\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}

Negative vector

-\overrightarrow{AB}=\overrightarrow{BA}

Combined transform

Apply right-hand transform first

Worked example 7.C — Vector path

$O$ is the origin; $\overrightarrow{OA}=\mathbf{a}$ , $\overrightarrow{OB}=\mathbf{b}$ . $M$ is the midpoint of $AB$ . Find $\overrightarrow{OM}$ .

\overrightarrow{OM}=\overrightarrow{OA}+\tfrac{1}{2}\overrightarrow{AB}=\mathbf{a}+\tfrac{1}{2}(\mathbf{b}-\mathbf{a})=\tfrac{1}{2}(\mathbf{a}+\mathbf{b}).

Find the magnitude of

\mathbf{v}=\binom{-5}{12}

and hence find a unit vector in the direction of

\mathbf{v}

|\mathbf{v}|=\sqrt{25+144}=\sqrt{169}=13

. Unit vector

=\dfrac{1}{13}\binom{-5}{12}

Enlargement with fractional scale factor

k = 0.5

: all linear dimensions are halved, and the area is scaled by

k^2 = 0.25

⚠️ Cambridge 0580 Traps — Transformations & Vectors

Single Transformation Rule: Examination instructions strictly forbid giving two transformations to describe a mapping. Stating 'reflection then translation' scores 0 marks even if both descriptions are correct. You must identify the single transformation: Translation, Reflection, Rotation, or Enlargement.

Required Transformation Details:

Translation: state column vector $\binom{x}{y}$ .
Reflection: state equation of mirror line (e.g. $y = x$ or $x = 2$ ).
Rotation: state angle (clockwise/anticlockwise) AND centre of rotation $(h, k)$ .
Enlargement: state scale factor $k$ AND centre of enlargement $(h, k)$ .

Negative Enlargement: Scale factor $k = -2$ produces an inverted shape on the opposite side of the centre of enlargement at twice the distance.

In triangle

OAB

\overrightarrow{OA} = 3\mathbf{a}

and

\overrightarrow{OB} = 6\mathbf{b}

. Point

P

lies on

AB

such that

AP : PB = 1 : 2

. Find

\overrightarrow{OP}

in terms of

\mathbf{a}

and

\mathbf{b}

in its simplest form.

Step 1: Express $\overrightarrow{AB}$ :

\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -3\mathbf{a} + 6\mathbf{b}

Step 2: Find $\overrightarrow{AP}$ :

\overrightarrow{AP} = \frac{1}{3}\overrightarrow{AB} = \frac{1}{3}(-3\mathbf{a} + 6\mathbf{b}) = -\mathbf{a} + 2\mathbf{b}

Step 3: Find $\overrightarrow{OP}$ :

\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = 3\mathbf{a} + (-\mathbf{a} + 2\mathbf{b}) = 2\mathbf{a} + 2\mathbf{b}

📝 Chapter Quiz

Questions Level

Probability & Statistics

📌 0580 Topic 8 — Probability & Statistics C E · Probability & tree diagrams · Conditional probability E · Mean, median, mode · Frequency diagrams & histograms E · Cumulative frequency & box plots E

Measuring chance and summarising data: the probability scale, combined events with tree diagrams, the three averages, and reading frequency and cumulative-frequency diagrams.

● Definition 8.1 — Probability

Probability runs from $0$ (impossible) to $1$ (certain). For equally likely outcomes, $P(E)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ .

Single event

P(E)=\dfrac{\text{favourable}}{\text{total}}

Complement

P(\text{not }E)=1-P(E)

AND (independent)

P(A\cap B)=P(A)\,P(B)

OR (exclusive)

P(A\cup B)=P(A)+P(B)

Worked example 8.A

A bag holds 3 red and 5 blue counters. One is drawn at random. Find $P(\text{red})$ .

There are $3+5=8$ equally likely counters, 3 of them red:

P(\text{red})=\frac{3}{8}.

Worked example 8.B

Find the mean of $4,\,8,\,10,\,6,\,2$ .

\bar{x}=\frac{4+8+10+6+2}{5}=\frac{30}{5}=6.

● Histograms (Frequency Density) E

For grouped data with unequal class widths, plot frequency density on the $y$ -axis, not frequency. $\text{Frequency density}=\dfrac{\text{frequency}}{\text{class width}}$ . Area of each bar equals the frequency for that class.

Frequency density

\dfrac{\text{frequency}}{\text{class width}}

Frequency from bar

area = class width × freq. density

Cumulative freq.

Running total of frequencies

IQR from box plot

Q_3-Q_1

Worked example 8.C — Histogram

A class interval 20–30 has frequency 15; class width = 10.

\text{Freq. density}=\frac{15}{10}=1.5.

Conversely, a bar of height 2.4 over interval 30–45 (width 15) represents frequency $2.4\times15=36$ .

● Conditional Probability & Tree Diagrams E

Without replacement: after removing one item, the total decreases, so probabilities on the second branch change. Conditional probability: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ . Multiply along branches for "and"; add separate branches for "or".

Worked example 8.D — Tree diagram without replacement

A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find $P(\text{both red})$ .

P(\text{both red})=\frac{5}{8}\times\frac{4}{7}=\frac{20}{56}=\frac{5}{14}.

Using the same bag (5 red, 3 blue), find the probability of drawing exactly one red ball.

P(\text{RB})+P(\text{BR})=\dfrac{5}{8}\cdot\dfrac{3}{7}+\dfrac{3}{8}\cdot\dfrac{5}{7}=\dfrac{15}{56}+\dfrac{15}{56}=\dfrac{30}{56}=\dfrac{15}{28}

Box-and-whisker plot displaying the five-number summary: Min = 15,

Q_1 = 25

, Median = 38,

Q_3 = 52

, Max = 70, with interquartile range

\text{IQR} = 27

⚠️ Cambridge 0580 Traps — Statistics & Probability

Histogram Frequency Density: In Cambridge IGCSE histograms, class widths are usually unequal. The vertical axis is ALWAYS Frequency Density, where $\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}$ . Frequency is given by the AREA of the bar, not its height!

Probability Without Replacement: When selecting items without replacement, reduce both the numerator and the denominator for the second draw: $\frac{4}{10} \times \frac{3}{9}$ . Forgetting to reduce the denominator to 9 is the classic exam error.

Estimating the Mean from Grouped Data: Always use the midpoint of each class interval, multiply by frequency, sum $\sum fx$ , and divide by total frequency $\sum f$ (never divide by the number of intervals).

The table shows the time,

t

minutes, taken by 50 students to complete a homework task.
Intervals:

0 < t \le 10

(freq: 8),

10 < t \le 20

(freq: 14),

20 < t \le 30

(freq: 18),

30 < t \le 40

(freq: 10).
Calculate an estimate of the mean time taken.

Find class midpoints ( $x$ ) and products ( $fx$ ):

Midpoint 5: $8 \times 5 = 40$
Midpoint 15: $14 \times 15 = 210$
Midpoint 25: $18 \times 25 = 450$
Midpoint 35: $10 \times 35 = 350$

\sum fx = 40 + 210 + 450 + 350 = 1050

\text{Estimated Mean} = \frac{\sum fx}{\sum f} = \frac{1050}{50} = 21\text{ minutes}

📝 Chapter Quiz

Questions Level

Functions

📌 0580 Topic 2 (Extended) — Functions E · Function notation · Domain & range · Composite functions · Inverse functions

Function notation, composite and inverse functions form a core Extended-tier topic.

▲ Reference — Functions

$f(x)$ gives the output for an input $x$ . The composite $fg(x)=f(g(x))$ is applied inside-out; the inverse $f^{-1}$ reverses $f$ .

Notation

f(x)

Composite

fg(x)

Inverse

f^{-1}(x)

Domain

valid inputs

Worked example 9.A

If $f(x)=2x+3$ , find $f^{-1}(x)$ .

f^{-1}(x)=\frac{x-3}{2}.

Worked example 9.B — Composite and inverse

Given $f(x)=2x+3$ and $g(x)=x^{2}$ , find (a) $fg(3)$ , (b) $f^{-1}(x)$ .

(a) $fg(3)=f(9)=2(9)+3=21$ .

(b) Let $y=2x+3$ . Swap and solve: $x=\dfrac{y-3}{2}$ . So $f^{-1}(x)=\dfrac{x-3}{2}$ .

h(x)=\dfrac{3x-1}{2}

. Find

h^{-1}(4)

y=\dfrac{3x-1}{2}\Rightarrow2y=3x-1\Rightarrow x=\dfrac{2y+1}{3}

. So

h^{-1}(x)=\dfrac{2x+1}{3}

and

h^{-1}(4)=\dfrac{9}{3}=3

Linear function

f(x) = 2x + 3

with gradient 2 and y-intercept

(0, 3)

⚠️ Cambridge 0580 Traps — Function Notation & Inverses

Inverse vs. Reciprocal: $f^{-1}(x)$ denotes the inverse function, NOT the reciprocal $\frac{1}{f(x)}$ . For example, if $f(x) = 3x - 1$ , $f^{-1}(x) = \frac{x+1}{3}$ , NOT $\frac{1}{3x-1}$ .

Composite Function Order: The notation $fg(x)$ means $f(g(x))$ — evaluate $g(x)$ first, then substitute that result into $f$ . Applying them in reverse order $gf(x)$ yields a completely different function.

Self-Inverse Functions: The graph of $f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ . Reciprocal functions $f(x) = \frac{k}{x}$ are self-inverse: $f^{-1}(x) = f(x)$ .

Worked example 9.C — Solving an equation with composite functions

Given $f(x) = 3x - 2$ and $g(x) = \frac{10}{x+1}$ , solve $fg(x) = 4$ .

Step 1: Express $fg(x)$ :

fg(x) = f\left(\frac{10}{x+1}\right) = 3\left(\frac{10}{x+1}\right) - 2 = \frac{30}{x+1} - 2

Step 2: Set equal to 4 and solve:

\frac{30}{x+1} - 2 = 4 \implies \frac{30}{x+1} = 6 \implies 30 = 6(x+1) \implies x+1 = 5 \implies x = 4

f(x) = 4x - 3

and

g(x) = x^2 + 1

, find (a)

gf(2)

and (b) an expression for

g^{-1}(x)

for

x \ge 1

(a) Inner function first: $f(2) = 4(2) - 3 = 5$ . Then $g(5) = 5^2 + 1 = 26$ .

(b) Inverse of $g$ : Let $y = x^2 + 1 \implies x^2 = y - 1 \implies x = \sqrt{y - 1}$ . Thus $g^{-1}(x) = \sqrt{x - 1}$ for $x \ge 1$ .

📝 Chapter Quiz

Questions Level

Sets & Venn Diagrams

📌 0580 Topic 1/2 — Sets & Venn Diagrams C E · Set notation · Union, intersection, complement · Two- and three-set Venn diagrams · Solving problems with $n(A)$

Sets group elements; Venn diagrams display unions, intersections and complements.

▲ Reference — Set notation

Union $A\cup B$ (in either set), intersection $A\cap B$ (in both), complement $A'$ (not in $A$ ); $n(A)$ counts the elements of $A$ .

Union

A\cup B

Intersection

A\cap B

Complement

A'

Subset

A\subseteq B

Worked example 10.A

If $A=\{1,2,3\}$ and $B=\{2,3,4\}$ , find $A\cap B$ .

\{2,\,3\}.

Worked example 10.B — Three-set Venn diagram

In a class of 30: 14 play football (F), 11 play basketball (B), 8 play tennis (T); 4 play F and B, 3 play F and T, 2 play B and T, 1 plays all three.

Use the formula: $n(F\cup B\cup T)=n(F)+n(B)+n(T)-n(F\cap B)-n(F\cap T)-n(B\cap T)+n(F\cap B\cap T)$

=14+11+8-4-3-2+1=25.

So $30-25=5$ students play none of the three sports.

\mathcal{E}=\{1,2,3,4,5,6,7,8,9,10\}

A=\{2,4,6,8\}

B=\{3,6,9\}

. List

A\cap B

A\cup B

, and

A'

A\cap B=\{6\}

A\cup B=\{2,3,4,6,8,9\}

A'=\{1,3,5,7,9,10\}

Venn diagram representing two intersecting sets

A

and

B

partitioned into disjoint regions:

A \cap B'

A \cap B

A' \cap B

, and exterior

(A \cup B)'

⚠️ Cambridge 0580 Traps — Set Notation & Venn Diagrams

Union vs. Intersection: $A \cup B$ is the UNION (all elements in $A$ OR $B$ OR both). $A \cap B$ is the INTERSECTION (only elements common to BOTH $A$ and $B$ ).

Complement De Morgan’s Laws: $(A \cup B)' = A' \cap B'$ (elements outside both sets). Contrast this with $A' \cup B' = (A \cap B)'$ (everything except the central intersection).

Number of Elements vs. The Set Itself: $n(A)$ is a number (integer count). Never write $n(A) = \{2, 3\}$ . The set itself is $A = \{2, 3\}$ , and its count is $n(A) = 2$ .

Worked example 10.C — Venn diagram with an unknown x

In a club of 40 members, 25 play badminton (B) and 20 play squash (S). 6 members play neither sport. Find how many play both sports.

Let $x = n(B \cap S)$ . The number who play badminton only is $25 - x$ , squash only is $20 - x$ , both is $x$ , and neither is 6.

(25 - x) + x + (20 - x) + 6 = 40

51 - x = 40 \implies x = 11

Therefore, 11 members play both sports.

In a survey of 60 students, 35 study French, 28 study German, and 12 study both languages. Find (a) the number of students who study neither language, and (b) the probability that a student chosen at random studies French given that they study German.

(a) Inclusion-Exclusion:

n(F \cup G) = n(F) + n(G) - n(F \cap G) = 35 + 28 - 12 = 51

n(\text{neither}) = 60 - 51 = 9\text{ students}

(b) Conditional probability:

P(F \mid G) = \frac{n(F \cap G)}{n(G)} = \frac{12}{28} = \frac{3}{7} \approx 0.429

📝 Chapter Quiz

Questions Level

Practice set

Fifty exam-style questions spanning all eight topic areas — Number through Probability & Statistics — and tagged by difficulty. Click any problem to reveal a full worked solution.

Write $0.000452$ in standard form.

Basic

4.52\times 10^{-4}.

Evaluate $\dfrac{3}{4}+\dfrac{1}{6}$ .

Basic

Common denominator 12:

\frac{9}{12}+\frac{2}{12}=\frac{11}{12}.

Increase 80 by 15%.

Basic

80\times 1.15=92.

A jacket costs \$60 after a 20% discount. Find its original price.

Intermediate

\$60 is 80% of the original:

\frac{60}{0.8}=\$75.

Share \$120 in the ratio $2:3$ .

Intermediate

5 parts, each $120\div 5=24$ :

\$48 \ \text{and}\ \$72.

Evaluate $2^{5}\times 2^{3}$ .

Basic

2^{5+3}=2^{8}=256.

Write 56 as a product of its prime factors.

Intermediate

56=2^{3}\times 7.

A length is 4.7 cm, correct to 1 decimal place. Write its lower bound.

Intermediate

Half of $0.1$ below the rounded value:

4.65\ \text{cm}.

Expand $3(2x-5)$ .

Basic

6x-15.

Solve $4x+7=23$ .

Basic

4x=16,\qquad x=4.

Factorise $x^{2}+5x+6$ .

Intermediate

Two numbers multiplying to 6, adding to 5:

(x+2)(x+3).

Solve $x^{2}-7x+10=0$ .

Intermediate

$(x-2)(x-5)=0$ :

x=2\ \text{or}\ x=5.

Make $r$ the subject of $C=2\pi r$ .

Intermediate

r=\frac{C}{2\pi}.

Solve simultaneously: $x+y=7$ , $2x-y=2$ .

Intermediate

Add: $3x=9 \Rightarrow x=3$ , then $y=4$ :

x=3,\ y=4.

Find the $n$ th term of $5,\,8,\,11,\,14,\ldots$

Intermediate

3n+2.

Solve the inequality $2x-3<9$ .

Advanced

2x<12,\qquad x<6.

Find the gradient of the line through $(1,2)$ and $(4,11)$ .

Basic

m=\frac{11-2}{4-1}=3.

Find the midpoint of $(2,3)$ and $(8,7)$ .

Intermediate

\Big(\tfrac{2+8}{2},\tfrac{3+7}{2}\Big)=(5,5).

A line has gradient 2 and passes through $(0,-1)$ . Write its equation.

Intermediate

y=2x-1.

Find the length of the segment from $(1,2)$ to $(4,6)$ .

Advanced

\sqrt{3^{2}+4^{2}}=\sqrt{25}=5.

Find the area of a circle of radius 7. Take $\pi=\tfrac{22}{7}$ .

Basic

\pi r^{2}=\frac{22}{7}\times 49=154.

Find the circumference of a circle of radius 5. Take $\pi=3.14$ .

Basic

2\pi r=2\times 3.14\times 5=31.4.

Find the area of a triangle with base 10 and height 6.

Intermediate

\tfrac12\times 10\times 6=30.

Find the volume of a cuboid measuring $4\times 5\times 3$ .

Intermediate

4\times 5\times 3=60.

Find the volume of a cylinder with $r=3$ , $h=10$ . Take $\pi=3.14$ .

Intermediate

\pi r^{2}h=3.14\times 9\times 10=282.6.

Find the volume of a sphere of radius 3. Take $\pi=3.14$ .

Advanced

\tfrac{4}{3}\pi r^{3}=\tfrac{4}{3}\times 3.14\times 27=113.04.

Find the area of a trapezium with parallel sides 6 and 10 and height 4.

Intermediate

\tfrac12(6+10)\times 4=32.

Find the total surface area of a cube of side 5.

Advanced

Six square faces:

6\times 5^{2}=150.

Two angles of a triangle are $50^{\circ}$ and $60^{\circ}$ . Find the third.

Basic

180^{\circ}-50^{\circ}-60^{\circ}=70^{\circ}.

Find the sum of the interior angles of a hexagon.

Intermediate

(6-2)\times 180^{\circ}=720^{\circ}.

Find each interior angle of a regular pentagon.

Intermediate

\frac{(5-2)\times 180^{\circ}}{5}=108^{\circ}.

A regular polygon has an exterior angle of $30^{\circ}$ . How many sides has it?

Intermediate

\frac{360^{\circ}}{30^{\circ}}=12\ \text{sides}.

State the size of the angle in a semicircle.

Intermediate

90^{\circ}.

$A$ and $B$ lie on a circle, centre $O$ , with $\angle AOB=80^{\circ}$ . Find the angle subtended by $AB$ at the circumference.

Advanced

Inscribed angle is half the central angle:

\tfrac12\times 80^{\circ}=40^{\circ}.

A right triangle has the two shorter sides 3 and 4. Find the hypotenuse.

Basic

\sqrt{3^{2}+4^{2}}=5.

In a right triangle the side opposite $\theta$ is 5 and the hypotenuse is 10. Find $\theta$ .

Intermediate

$\sin\theta=\tfrac{5}{10}=0.5$ :

\theta=30^{\circ}.

A right triangle has hypotenuse 10 and an angle of $30^{\circ}$ . Find the opposite side.

Intermediate

10\sin 30^{\circ}=5.

A triangle has $a=7$ , $b=8$ and included angle $C=60^{\circ}$ . Find $c$ .

Advanced

Cosine rule, $\cos 60^{\circ}=0.5$ :

c^{2}=49+64-56=57,\quad c=\sqrt{57}.

Find the hypotenuse of a right triangle with legs 6 and 8.

Intermediate

\sqrt{6^{2}+8^{2}}=10.

In triangle $ABC$ , $a=10$ , $A=30^{\circ}$ , $B=45^{\circ}$ . Find $b$ (sine rule).

Advanced

b=\frac{a\sin B}{\sin A}=\frac{10\sin 45^{\circ}}{\sin 30^{\circ}}=10\sqrt2\approx 14.14.

Reflect the point $(3,2)$ in the $x$ -axis.

Basic

(3,-2).

Rotate the point $(1,0)$ by $90^{\circ}$ anticlockwise about the origin.

Intermediate

$(x,y)\to(-y,x)$ :

(0,1).

Given $\mathbf{a}=\begin{pmatrix}2\\3\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}1\\-1\end{pmatrix}$ , find $\mathbf{a}+\mathbf{b}$ .

Intermediate

\begin{pmatrix}3\\2\end{pmatrix}.

Find the magnitude of $\begin{pmatrix}3\\4\end{pmatrix}$ .

Advanced

\sqrt{3^{2}+4^{2}}=5.

A fair die is rolled. Find $P(\text{even number})$ .

Basic

Three of six outcomes are even:

\frac{3}{6}=\frac12.

Find the mean of $4,\,8,\,10,\,6,\,2$ .

Intermediate

\frac{30}{5}=6.

Find the median of $3,\,7,\,2,\,9,\,5$ .

Intermediate

Ordered: $2,3,5,7,9$ ; middle value:

5.

A bag holds 3 red and 5 blue counters. Find $P(\text{red})$ .

Intermediate

\frac{3}{8}.

Find the mode of $2,\,3,\,3,\,5,\,7,\,3,\,8$ .

Intermediate

The most frequent value:

3.

Two fair coins are tossed. Find $P(\text{two heads})$ .

Advanced

\tfrac12\times\tfrac12=\frac14.

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