Geometry | GCSE - Wyatt's Notes

----------------------- | -------------------- | | Angles on a straight line | Sum to | | Angles at a point | Sum to | |” date: 2026-04-14 tags:

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Geometry is the mathematics of shape, size, and space. The key insight is that geometric properties (angles, lengths, areas) follow logical rules that can be proved from a small number of axioms. Understanding these rules lets you calculate unknown measurements from known ones.

Why angles matter: Angles describe rotation and direction. Angles on a straight line sum to 180°, angles around a point sum to 360°, and angles in a triangle sum to 180°. These rules are not arbitrary — they follow from the geometry of flat space. In curved space (like on a sphere), triangle angles sum to more than 180°.

Congruence vs similarity intuition: Congruent shapes are identical in size and shape — they can be mapped onto each other by translation, rotation, or reflection. Similar shapes have the same shape but different sizes — one is a scaled version of the other. Congruence preserves lengths and angles; similarity preserves angles and ratios of lengths.

Info: Board Coverage AQA Paper 1 & 2 | Edexcel Paper 1 & 2 | OCR Paper 2 & 3 | WJEC Unit 2

Theorem (Angle sum of a triangle). The interior angles of any triangle sum to $180^{\circ}$.

Proof. Let triangle $\triangle ABC$ have vertices $A$, $B$, $C$. Draw a line through $C$ Parallel to $AB$. Label the intersection points of this line with the exterior of the triangle as $D$ and $E$ (so $D$ is on the side of $B$ and $E$ is on the side of $A$).

By alternate angles: $\angle B = \angle BCD$ and $\angle A = \angle ACE$.

Since $D$, $C$, $E$ lie on a straight line: $\angle BCD + \angle ACB + \angle ACE = 180^{\circ}$.

Therefore $\angle A + \angle B + \angle C = 180^{\circ}$. $\blacksquare$

Theorem (Angle sum of a quadrilateral). The interior angles of any quadrilateral sum to $360^{\circ}$.

Proof. Draw a diagonal, dividing the quadrilateral into two triangles. Each triangle has angles Summing to $180^{\circ}$ So the total is $360^{\circ}$. $\blacksquare$

Corollary. The interior angles of any $n$-sided polygon sum to $180(n - 2)^{\circ}$.

Proof by induction. A triangle ($n = 3$) has sum $180^{\circ}$. Adding a vertex to an $(n-1)$-gon creates a triangle, adding $180^{\circ}$. So an $n$-gon has sum $180 + 180(n - 3) = 180(n - 2)^{\circ}$. $\blacksquare$

Worked Example. In the diagram, line $AB$ is parallel to line $CD$. A transversal intersects $AB$ at $E$ and $CD$ at $F$. If $\angle AEF = 65^{\circ}$Find all the other angles at $E$ and $F$.

$\angle AEF = \angle EFD = 65^{\circ}$ (alternate angles).

$\angle BEF = 180^{\circ} - 65^{\circ} = 115^{\circ}$ (angles on a straight line).

$\angle EFC = 115^{\circ}$ (corresponding to $\angle BEF$).

$\angle CFD = 65^{\circ}$ (vertically opposite $\angle EFD$).

$\angle EFB = \angle EFD = 65^{\circ}$ (vertically opposite).

Worked Example (Higher Tier). Two parallel lines are cut by a transversal. One interior angle is $3x + 10^{\circ}$ and the co-interior angle is $2x + 20^{\circ}$. Find the value of $x$.

Since co-interior angles sum to $180^{\circ}$:

$(3x + 10) + (2x + 20) = 180$

$5x + 30 = 180$

$5x = 150$

$x = 30$

The sum of interior angles of an $n$-sided polygon:

$S = 180(n - 2)^{\circ}$

The interior angle of a regular $n$-sided polygon:

\mathrm{Each interior angle = \frac{180(n - 2)}{n}^{\circ}

The exterior angle of a regular polygon:

\mathrm{Each exterior angle = \frac{360}{n}^{\circ}

Worked Example. Find the interior angle of a regular pentagon.

\mathrm{Interior angle = \frac{180(5 - 2)}{5} = \frac{540}{5} = 108^{\circ}

Worked Example (Higher Tier). A regular polygon has an interior angle of $150^{\circ}$. How many Sides does it have?

$\frac{180(n - 2)}{n} = 150$ $180n - 360 = 150n$ $30n = 360$ $n = 12$

It is a regular dodecagon (12 sides).

Worked Example (Higher Tier). Find the sum of the interior angles of a polygon with 15 sides.

$S = 180(15 - 2) = 180 \times 13 = 2340^{\circ}$

Worked Example (Higher Tier). Find the exterior angle of a regular decagon, and hence find the Interior angle.

\mathrm{Exterior angle = \frac{360}{10} = 36^{\circ}

\mathrm{Interior angle = 180 - 36 = 144^{\circ}

A bearing is an angle measured clockwise from north, always given as a three-figure number (e.g. $045^{\circ}$Not $45^{\circ}$).

Worked Example. A ship sails from port $A$ on a bearing of $070^{\circ}$ for 80 km to point $B$. It then sails on a bearing of $150^{\circ}$ for 60 km to point $C$. Find the bearing of $C$ from $A$.

The angle $\angle NAB = 70^{\circ}$ and $\angle NBC = 150^{\circ}$.

The angle between $AB$ and north at $B$ is $180^{\circ} - 70^{\circ} = 110^{\circ}$ (back bearing).

The angle $\angle ABC = 110^{\circ} + 150^{\circ} = 260^{\circ}$… This is the external angle. The Internal angle is $360^{\circ} - 260^{\circ} = 100^{\circ}$.

Using the cosine rule on $\triangle ABC$:

$AC^2 = 80^2 + 60^2 - 2 \times 80 \times 60 \times \cos(100^{\circ})$ $AC^2 = 6400 + 3600 - 9600 \times (-0.1736\ldots)$ $AC^2 = 10000 + 1667.1\ldots = 11667.1\ldots$ AC \approx 108.0 \mathrm{ km

Using the sine rule to find $\angle BAC$:

$\frac{\sin \angle BAC}{60} = \frac{\sin 100^{\circ}}{108.0}$ $\sin \angle BAC = \frac{60 \times 0.9848}{108.0} = 0.5471\ldots$ $\angle BAC = 33.2^{\circ}$

Bearing of $C$ from $A = 70^{\circ} + 33.2^{\circ} = 103.2^{\circ} \approx 103^{\circ}$.

Theorem. In a right-angled triangle with hypotenuse $c$ and legs $a$ and $b$:

$a^2 + b^2 = c^2$

Proof (area-based). Consider a square of side $(a + b)$. Place four identical right-angled Triangles inside, each with legs $a$ and $b$Arranged so that their hypotenuses form a smaller Square of side $c$ in the centre.

The area of the large square equals the area of the four triangles plus the area of the inner Square:

$(a + b)^2 = 4 \times \frac{1}{2}ab + c^2$ $a^2 + 2ab + b^2 = 2ab + c^2$ $a^2 + b^2 = c^2 \quad \blacksquare$

Converse of Pythagoras’ Theorem. If $a^2 + b^2 = c^2$ for a triangle with sides $a, b, c$ where $c$ is the longest side, then the triangle is right-angled.

Worked Example. Is a triangle with sides 5 cm, 12 cm, and 13 cm right-angled?

$5^2 + 12^2 = 25 + 144 = 169 = 13^2$. Yes, it is right-angled.

Worked Example. A ladder of length 10 m leans against a wall with its foot 6 m from the base of The wall. How high up the wall does it reach?

$h^2 + 6^2 = 10^2$ $h^2 = 100 - 36 = 64$ h = 8 \mathrm{ m

Worked Example (Higher Tier). Is a triangle with sides 7 cm, 11 cm, and 13 cm acute, Right-angled, or obtuse?

$7^2 + 11^2 = 49 + 121 = 170$. Since $170 \gt 13^2 = 169$The triangle is acute (the angle opposite The longest side is less than $90^{\circ}$).

Test for triangle type:

For a right-angled triangle with angle $\theta$:

\sin \theta = \frac{\mathrm{opposite}{\mathrm{hypotenuse}, \quad \cos \theta = \frac{\mathrm{adjacent}{\mathrm{hypotenuse}, \quad \tan \theta = \frac{\mathrm{opposite}{\mathrm{adjacent}

Proof that $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

\tan \theta = \frac{\mathrm{opp}{\mathrm{adj} = \frac{\mathrm{opp/\mathrm{hyp}{\mathrm{adj/\mathrm{hyp} = \frac{\sin \theta}{\cos \theta} \quad \blacksquare

Proof that $\sin^2\theta + \cos^2\theta = 1$.

By Pythagoras: \mathrm{opp^2 + \mathrm{adj^2 = \mathrm{hyp^2.

Dividing by \mathrm{hyp^2:

\frac{\mathrm{opp^2}{\mathrm{hyp^2} + \frac{\mathrm{adj^2}{\mathrm{hyp^2} = 1

$\sin^2\theta + \cos^2\theta = 1 \quad \blacksquare$

Worked Example. Find the length of the hypotenuse in a right-angled triangle where the opposite Side is 5 cm and the angle is $35^{\circ}$.

$\sin 35^{\circ} = \frac{5}{h}$ h = \frac{5}{\sin 35^{\circ}} = \frac{5}{0.5736\ldots} = 8.72 \mathrm{ cm (to 3 s.f.)

Worked Example. Find the angle $\theta$ in a right-angled triangle where the adjacent side is 8 Cm and the hypotenuse is 15 cm.

$\cos \theta = \frac{8}{15}$ $\theta = \cos^{-1}\!\left(\frac{8}{15}\right) \approx 57.8^{\circ}$

For any triangle $\triangle ABC$:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Used when you know: an angle and its opposite side, plus one other angle or side.

Proof sketch. Drop an altitude from $C$ to $AB$. In the two right-angled triangles formed, Express the height as both $b \sin A$ and $a \sin B$. Setting equal: $b \sin A = a \sin B$Giving $\frac{a}{\sin A} = \frac{b}{\sin B}$. $\blacksquare$

Worked Example (Higher Tier). In $\triangle ABC$, $a = 10$ cm, $A = 45^{\circ}$ $B = 70^{\circ}$. Find $b$.

$\frac{b}{\sin 70^{\circ}} = \frac{10}{\sin 45^{\circ}}$ b = \frac{10 \sin 70^{\circ}}{\sin 45^{\circ}} = \frac{10 \times 0.9397}{0.7071} \approx 13.29 \mathrm{ cm

For any triangle $\triangle ABC$:

$a^2 = b^2 + c^2 - 2bc \cos A$

Rearranged to find an angle:

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$

Used when you know: two sides and the included angle (to find the third side), or all three sides (to find an angle).

Worked Example. In $\triangle ABC$, $a = 8$ cm, $b = 5$ cm, $c = 7$ cm. Find angle $A$.

$\cos A = \frac{25 + 49 - 64}{2 \times 5 \times 7} = \frac{10}{70} = \frac{1}{7}$ A = \cos^{-1}\!\left(\frac{1}{7}\right) = 81.8^{\circ} \mathrm{ (to 1 d.p.)

Worked Example (Higher Tier). In $\triangle ABC$, $a = 12$ cm, $b = 8$ cm, $C = 60^{\circ}$. Find $c$.

$c^2 = 144 + 64 - 2 \times 12 \times 8 \times \cos 60^{\circ}$ $c^2 = 208 - 96 = 112$ c = \sqrt{112} = 4\sqrt{7} \approx 10.58 \mathrm{ cm

\mathrm{Area = \frac{1}{2}ab \sin C

Where $a$ and $b$ are two sides and $C$ is the included angle.

Worked Example. Find the area of $\triangle ABC$ where $a = 10$ cm, $b = 8$ cm, and $C = 45^{\circ}$.

\mathrm{Area = \frac{1}{2} \times 10 \times 8 \times \sin 45^{\circ} = 40 \times \frac{\sqrt{2}}{2} = 20\sqrt{2} \approx 28.3 \mathrm{ cm^2

When using the sine rule to find an angle, there may be two possible solutions: $\theta$ and $180^{\circ} - \theta$.

Worked Example. In $\triangle ABC$, $a = 8$ cm, $b = 10$ cm, $A = 40^{\circ}$. Find angle $B$.

$\frac{\sin B}{10} = \frac{\sin 40^{\circ}}{8}$ $\sin B = \frac{10 \sin 40^{\circ}}{8} = \frac{10 \times 0.6428}{8} = 0.8035$

$B = \sin^{-1}(0.8035) \approx 53.5^{\circ}$ or $B = 180^{\circ} - 53.5^{\circ} = 126.5^{\circ}$.

Both are valid since $53.5 + 40 = 93.5 \lt 180$ and $126.5 + 40 = 166.5 \lt 180$.

When to check for the ambiguous case: Only when finding an angle using the sine rule. If Finding a side, there is at most one solution.

Proof. Let $O$ be the centre of a circle. Let arc $AB$ subtend angle $\angle AOB = 2\theta$ at The centre and angle $\angle ACB = \theta$ at a point $C$ on the circumference.

Draw the radius $OC$. Since $OA = OC$ (radii), $\triangle OAC$ is isosceles, so $\angle OAC = \angle OCA$.

Similarly, $OB = OC$ So $\triangle OBC$ is isosceles, and $\angle OBC = \angle OCB$.

The exterior angle at $O$ for $\triangle OAC$: $2\angle OCA = \angle AOC$.

The exterior angle at $O$ for $\triangle OBC$: $2\angle OCB = \angle BOC$.

Adding: $\angle AOC + \angle BOC = 2\angle OCA + 2\angle OCB$

$\angle AOB = 2(\angle OCA + \angle OCB) = 2\angle ACB \quad \blacksquare$

Proof. Let $AB$ be a diameter of a circle with centre $O$. Let $C$ be any point on the Circumference.

Since $AB$ is a diameter, $\angle AOB = 180^{\circ}$.

By the angle at centre theorem: $\angle AOB = 2\angle ACB$.

Therefore $\angle ACB = 90^{\circ}$. $\blacksquare$

Proof. Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$. The arc $ABC$ subtends angle $\angle ADC$ at the circumference and angle $\angle AOC$ at the centre. Similarly, arc $ADC$ Subtends $\angle ABC$ and $\angle AOD$.

The arcs $ABC$ and $ADC$ together make the full circle, so $\angle AOC + \angle AOD = 360^{\circ}$.

By the angle at centre theorem: $\angle ADC = \frac{1}{2}\angle AOC$ and $\angle ABC = \frac{1}{2}\angle AOD$.

Adding: $\angle ABC + \angle ADC = \frac{1}{2} \times 360^{\circ} = 180^{\circ}$. $\blacksquare$

Worked Example. $A$, $B$, $C$ And $D$ lie on a circle. $\angle ABC = 75^{\circ}$ and $\angle CAD = 40^{\circ}$. Find $\angle ABD$.

$\angle ABC$ and $\angle ADC$ are opposite angles of cyclic quadrilateral $ABCD$: $\angle ADC = 180^{\circ} - 75^{\circ} = 105^{\circ}$.

$\angle CAD$ and $\angle CBD$ are in the same segment (subtended by arc $CD$), so $\angle CAD = \angle CBD = 40^{\circ}$.

Therefore $\angle ABD = \angle ABC - \angle CBD = 75^{\circ} - 40^{\circ} = 35^{\circ}$.

Worked Example (Higher Tier). $AB$ is a diameter of a circle with centre $O$. $C$ is a point on The circle such that $\angle BAC = 32^{\circ}$. Find $\angle OCA$.

Since $AB$ is a diameter, $\angle ACB = 90^{\circ}$.

$\angle ABC = 90^{\circ} - 32^{\circ} = 58^{\circ}$.

Since $OA = OC$ (radii), $\triangle OAC$ is isosceles: $\angle OCA = \angle OAC = 32^{\circ}$.

Worked Example (Higher Tier). $AB$ and $AC$ are tangents to a circle at points $B$ and $C$ Respectively. Prove that $AB = AC$.

Join $O$ to $A$, $B$ And $C$. Since $OB$ and $OC$ are radii, and tangents are perpendicular to Radii at the point of contact, $\angle OBA = \angle OCA = 90^{\circ}$.

$OA$ is common, and $OB = OC$ (radii). By RHS (right angle, hypotenuse, side), $\triangle OBA \cong \triangle OCA$.

Therefore $AB = AC$. $\blacksquare$

Proof of the area of a trapezium. A trapezium with parallel sides $a$ and $b$ and height $h$ can Be divided into a rectangle and two triangles. The rectangle has area $ah$ and the two triangles Have total area $\frac{1}{2}(b-a)h + \frac{1}{2}(b-a)h = (b-a)h$. Total: $ah + (b-a)h = bh - ah + ah = (a+b)h/2$. $\blacksquare$

Worked Example. A cylinder has radius 5 cm and height 12 cm. Find its volume and total surface Area.

V = \pi \times 5^2 \times 12 = 300\pi \approx 942 \mathrm{ cm^3

\mathrm{SA = 2\pi \times 25 + 2\pi \times 5 \times 12 = 50\pi + 120\pi = 170\pi \approx 534 \mathrm{ cm^2

Worked Example (Higher Tier). A cone has base radius 6 cm and slant height 10 cm. Find its Volume.

The height $h$: $h^2 + 6^2 = 10^2$ So $h = 8$ cm.

V = \frac{1}{3}\pi \times 36 \times 8 = 96\pi \approx 301.6 \mathrm{ cm^3

Worked Example (Higher Tier). A solid hemisphere has radius 7 cm. Find its total surface area.

Curved surface area: $2\pi r^2 = 2\pi \times 49 = 98\pi$.

Flat face: $\pi r^2 = 49\pi$.

Total: 147\pi \approx 461.8 \mathrm{ cm^2.

Worked Example (Higher Tier). A frustum is formed by removing a small cone of height 4 cm from The top of a cone of height 12 cm. Both cones have the same base radius 5 cm. Find the volume of the Frustum.

The large cone has volume $\frac{1}{3}\pi \times 25 \times 12 = 100\pi$.

The small cone has height 4 cm. By similar triangles, the radius ratio is $4/12 = 1/3$ So the small Cone has radius $5/3$ cm.

Volume of small cone: $\frac{1}{3}\pi \times \frac{25}{9} \times 4 = \frac{100\pi}{27}$.

Volume of frustum: 100\pi - \frac{100\pi}{27} = \frac{2600\pi}{27} \approx 302.3 \mathrm{ cm^3.

Worked Example. Describe fully the transformation that maps $\triangle ABC$ with vertices at (1, 2)$$(3, 5)$$(5, 2) to $\triangle A'B'C'$ with vertices at (-1, -2)$$(-3, -5) $(-5, -2)$.

$(1, 2) \to (-1, -2)$: the $x$-coordinate is negated and the $y$-coordinate is negated. This is a Reflection in the origin, which is equivalent to a rotation of $180^{\circ}$ about the origin.

A vector has both magnitude and direction. We write vectors as column vectors or using bold letters.

Addition: $\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}$

Scalar multiplication: $k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}$

Magnitude: $\left|\begin{pmatrix} a \\ b \end{pmatrix}\right| = \sqrt{a^2 + b^2}$

Parallel vectors: $\mathbf{a}$ and $\mathbf{b}$ are parallel if $\mathbf{a} = k\mathbf{b}$ for Some scalar $k$.

Worked Example. Points A$$B And $C$ have position vectors \begin{pmatrix} 2 \\ 3 \end{pmatrix}$$\begin{pmatrix} 8 \\ 7 \end{pmatrix} And $\begin{pmatrix} 14 \\ 11 \end{pmatrix}$. Show that A$$B And $C$ are collinear.

$\overrightarrow{AB} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}, \qquad \overrightarrow{BC} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$

Since $\overrightarrow{AB} = \overrightarrow{BC}$The vectors are parallel and share point $B$ So $A$, $B$, $C$ are collinear. $\blacksquare$

The position vector of a point $P$ relative to an origin $O$ is $\overrightarrow{OP}$.

The vector from $A$ to $B$ is $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$.

The midpoint $M$ of $AB$ has position vector $\frac{1}{2}(\mathbf{a} + \mathbf{b})$.

Worked Example (Higher Tier). Point $P$ divides the line segment $AB$ in the ratio $2 : 3$. If $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} 11 \\ 9 \end{pmatrix}$Find $\overrightarrow{OP}$.

$\overrightarrow{OP} = \overrightarrow{OA} + \frac{2}{5}\overrightarrow{AB} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} + \frac{2}{5}\begin{pmatrix} 10 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} + \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 6 \end{pmatrix}$

Triangles are congruent if they are identical in shape and size. The conditions are:

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