Homeβ€Ί Mathematicsβ€Ί Unit 04 β€” Geometry & Measuresβ€Ί Trigonometry
Mathematics Β· AQA 8300 Β§G7

Trigonometry

Spec: AQA 8300 §G7 ⭐⭐⭐⭐ ⏱ 60 mins Boards: AQA · Edexcel · OCR Grade 9

Learning Objectives

  • Apply SOH-CAH-TOA to find sides and angles in right-angled triangles
  • Use exact trigonometric values for 30Β°, 45Β° and 60Β°
  • Apply the sine rule to non-right-angled triangles
  • Apply the cosine rule to find sides and angles
  • Use the formula Area = Β½ab sinC for the area of any triangle

πŸ”‘ Core Concepts

SOH-CAH-TOA in Right-Angled Triangles

Trigonometry in right-angled triangles links the angles to the ratios of the sides. The three side labels β€” Opposite (O), Adjacent (A) and Hypotenuse (H) β€” are always relative to the angle you are working with, not a fixed position in the triangle. The Hypotenuse is always the longest side, opposite the right angle.

πŸ“–
DEFINITION β€” SOH-CAH-TOA
For an acute angle $\theta$ in a right-angled triangle:
$\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$    $\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$    $\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}}$
SOH-CAH-TOA Formulae
$$\sin\theta = \frac{O}{H} \qquad \cos\theta = \frac{A}{H} \qquad \tan\theta = \frac{O}{A}$$
$O$ = Opposite side $A$ = Adjacent side $H$ = Hypotenuse $\theta$ = angle being used
🎯
EXAM TIP β€” Choosing the Right Ratio
Label all three sides relative to the angle first. Then identify which two sides are involved (one known, one unknown) and choose the ratio that uses exactly those two sides. If you have O and H, use sin; if A and H, use cos; if O and A, use tan.
🧠
MEMORY TRICK β€” SOH-CAH-TOA
"Some Old Hippos Can Always Hide Their Old Age" β€” the capitals give you S-O-H, C-A-H, T-O-A. Alternatively, write SOH-CAH-TOA as a triangle mnemonic on your exam paper immediately.

To find a side: rearrange the ratio formula so the unknown side is the subject. To find an angle: use the inverse function ($\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$).

Label sides O, A, H relative to angle
β†’
Identify which two sides are involved
β†’
Select sin / cos / tan
β†’
Rearrange and calculate
β†’
Check answer is reasonable

Exact Trigonometric Values

For the special angles 30Β°, 45Β° and 60Β°, trigonometric ratios take exact surd or fractional values. You must memorise these β€” exam questions often require exact answers rather than decimal approximations, and using exact values avoids rounding errors in multi-step problems.

πŸ“–
DEFINITION β€” Exact Values Table
Anglesincostan
0Β°$0$$1$$0$
30Β°$\dfrac{1}{2}$$\dfrac{\sqrt{3}}{2}$$\dfrac{1}{\sqrt{3}}$
45Β°$\dfrac{1}{\sqrt{2}}$$\dfrac{1}{\sqrt{2}}$$1$
60Β°$\dfrac{\sqrt{3}}{2}$$\dfrac{1}{2}$$\sqrt{3}$
90Β°$1$$0$undefined
🧠
MEMORY TRICK β€” Exact Values
For sin 0Β°, 30Β°, 45Β°, 60Β°, 90Β°, the values are $\sqrt{0}/2$, $\sqrt{1}/2$, $\sqrt{2}/2$, $\sqrt{3}/2$, $\sqrt{4}/2$ β€” i.e. $\sqrt{n}/2$ where $n$ = 0, 1, 2, 3, 4. For cos, read the sin column in reverse order. These come from the 30-60-90 triangle (sides 1, $\sqrt{3}$, 2) and the 45-45-90 triangle (sides 1, 1, $\sqrt{2}$).
βœ—
COMMON MISTAKE β€” tan 30Β° and tan 60Β°
Students often confuse $\tan 30Β° = \frac{1}{\sqrt{3}}$ and $\tan 60Β° = \sqrt{3}$. Note that $\tan 60Β°$ is the reciprocal of $\tan 30Β°$, and both are irrational. The rationalised form of $\tan 30Β°$ is $\frac{\sqrt{3}}{3}$.

The Sine Rule

The sine rule applies to any triangle (not just right-angled) and links each side to the sine of its opposite angle. It is used when you know two angles and a side (AAS or ASA) or two sides and a non-included angle (SSA β€” take care with the ambiguous case).

πŸ“–
DEFINITION β€” Sine Rule
In triangle $ABC$ with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$ respectively: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ Or, rearranged for finding angles: $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$
Sine Rule
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
$a, b, c$ = sides of the triangle $A, B, C$ = angles opposite to respective sides Use for finding sides: put the unknown side on top Use for finding angles: put the unknown angle on top
⚠️
IMPORTANT β€” Ambiguous Case (SSA)
When given two sides and a non-included angle, there may be two possible triangles (the ambiguous case). If $\sin B$ gives a value less than 1, then $B$ could be acute or obtuse. The obtuse solution is $B' = 180Β° - B$. Check whether both solutions are valid by verifying that all angles sum to 180Β° and all angles are positive. Full marks at Grade 9 require you to identify and present both solutions.
🎯
EXAM TIP β€” Sine Rule Setup
Only use two of the three fractions at a time. Always select the pair where you know (or want) both the numerator and denominator values. You do not need all three sides/angles β€” just two pairs.

The Cosine Rule

The cosine rule applies to any triangle and is used when you have SAS (two sides and the included angle) to find the third side, or SSS (all three sides) to find any angle. Unlike the sine rule, there is no ambiguous case.

πŸ“–
DEFINITION β€” Cosine Rule
To find a side when two sides and the included angle are known: $$a^2 = b^2 + c^2 - 2bc\cos A$$ Rearranged to find an angle when all three sides are known: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$
Cosine Rule
$$a^2 = b^2 + c^2 - 2bc\cos A$$
$a$ = side opposite angle $A$ $b, c$ = the other two sides $A$ = included angle between sides $b$ and $c$ For angles: rearrange to $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$
βœ—
COMMON MISTAKE β€” Cosine Rule Sign
The formula uses $-2bc\cos A$. When angle $A$ is obtuse, $\cos A$ is negative, so $-2bc\cos A$ becomes positive β€” meaning $a^2 > b^2 + c^2$. Students often forget this and get a smaller side than expected.
🎯
EXAM TIP β€” Sine Rule vs Cosine Rule
Use the sine rule when: you have AAS, ASA, or SSA (two angles + any side, or two sides + non-included angle).
Use the cosine rule when: you have SAS or SSS (two sides + included angle, or three sides).

Area of a Triangle β€” Β½ab sinC

For any triangle where the perpendicular height is unknown but two sides and the included angle are known, the area formula using trigonometry is far more efficient than $\frac{1}{2} \times \text{base} \times \text{height}$.

πŸ“–
DEFINITION β€” Trigonometric Area Formula
$$\text{Area} = \frac{1}{2}ab\sin C$$ where $a$ and $b$ are two sides of the triangle and $C$ is the angle between them (the included angle).
Area of a Triangle
$$\text{Area} = \tfrac{1}{2}\,ab\sin C$$
$a, b$ = two known sides $C$ = the angle between sides $a$ and $b$ Works for any triangle β€” right-angled or not
βœ—
COMMON MISTAKE β€” Included Angle
The angle $C$ must be the angle between the two sides $a$ and $b$ (the included angle). Using a non-included angle gives the wrong answer. Always label the triangle carefully before substituting.

3D Trigonometry

3D trigonometry problems require you to identify right-angled triangles within a 3D shape, apply Pythagoras' theorem to find lengths, and then apply trigonometry. The key skill is extracting a 2D triangle from a 3D diagram and working systematically.

πŸ“–
DEFINITION β€” 3D Trig Strategy
In a 3D problem (e.g. cuboid, pyramid, wedge):
  1. Identify the triangle containing the required side or angle.
  2. Find any missing lengths using Pythagoras' theorem on a different face or base.
  3. Substitute into SOH-CAH-TOA or the sine/cosine rule as appropriate.
  4. Always draw and label the 2D triangle you are working with.
🎯
EXAM TIP β€” Space Diagonal of a Cuboid
For a cuboid with length $l$, width $w$, height $h$: the space diagonal $d = \sqrt{l^2 + w^2 + h^2}$ (apply Pythagoras in two stages: base diagonal first, then include height). This is a very common 3D trig setting.

Bearings with Trigonometry

Bearings are measured clockwise from North, written as three digits (e.g. 045Β°, 270Β°). In bearing problems, the sine and cosine rules are applied to the triangle formed by the paths of travel. You must find the interior angles of the triangle from the given bearings.

πŸ“–
DEFINITION β€” Converting Bearings to Triangle Angles
To find the interior angle of a triangle from bearings:
  • Draw a North line at each vertex.
  • Use alternate angles (parallel North lines) and co-interior angles to find each interior angle.
  • Then apply the sine or cosine rule to find unknown sides or angles.
βœ—
COMMON MISTAKE β€” Bearing Direction
Always measure bearings clockwise from North. A bearing of 245Β° is not the same as South-West at 45Β° β€” the interior angle of the triangle formed will differ. Sketch a diagram with North arrows before calculating.

πŸ—ΊοΈ Visual Notes

Trigonometry
SOH-CAH-TOA
  • Right-angled triangles only
  • $\sin\theta = O/H$
  • $\cos\theta = A/H$
  • $\tan\theta = O/A$
Exact Values
  • sin 30Β° = Β½, cos 30Β° = √3/2
  • sin 45Β° = cos 45Β° = 1/√2
  • sin 60Β° = √3/2, cos 60Β° = Β½
  • tan 45Β° = 1, tan 60Β° = √3
Sine Rule
  • Any triangle
  • Use: AAS, ASA, SSA
  • $a/\sin A = b/\sin B$
  • Watch: ambiguous case (SSA)
Cosine Rule
  • Any triangle, no ambiguity
  • Use: SAS β†’ find side
  • Use: SSS β†’ find angle
  • $a^2 = b^2 + c^2 - 2bc\cos A$
Area Formula
  • Area = Β½ab sin C
  • C is the included angle
  • Works for all triangles
  • Link: combine with cosine rule
3D & Bearings
  • Extract 2D cross-sections
  • Apply Pythagoras first
  • Bearings: clockwise from N
  • Draw North arrows at each point

Formula Selection Guide

Given Information Want to Find Use Notes
Right-angled, 2 sidesAngleSOH-CAH-TOAUse inverse trig
Right-angled, angle + sideSideSOH-CAH-TOARearrange formula
AAS or ASASideSine RuleFind third angle first if AAS
SSA (two sides, non-included angle)Angle or sideSine RuleCheck ambiguous case!
SAS (two sides, included angle)Third sideCosine Rule$a^2 = b^2 + c^2 - 2bc\cos A$
SSS (all three sides)Any angleCosine Rule$\cos A = \frac{b^2+c^2-a^2}{2bc}$
Two sides + included angleAreaΒ½ab sin CC is the angle between a and b

Decision Tree β€” Which Formula?

Is the triangle right-angled?
β†’ YES β†’
Use SOH-CAH-TOA (or Pythagoras)
β†’ NO β†’
Do you have SAS or SSS?
β†’ YES β†’
Use Cosine Rule
β†’ NO β†’
Use Sine Rule (AAS/ASA/SSA)

Deriving Exact Values from Special Triangles

Triangle Type Angles Side lengths Values derived
Half equilateral triangle30°, 60°, 90°1, √3, 2sin/cos/tan 30° and 60°
Isosceles right triangle45°, 45°, 90°1, 1, √2sin/cos/tan 45°

✏️ Worked Examples

Grade 4–5 Β· SOH-CAH-TOA
In a right-angled triangle, the hypotenuse is 12 cm and one angle is 35Β°. Find the side opposite the 35Β° angle. Give your answer to 3 significant figures.
1
Label the sides
The hypotenuse (H) = 12 cm. The side we want is opposite (O) the 35Β° angle. We have O and H, so use sin.
2
Write the formula
$\sin\theta = \dfrac{O}{H}$, so $\sin 35Β° = \dfrac{O}{12}$
3
Rearrange and calculate
$O = 12 \times \sin 35Β° = 12 \times 0.5736... = 6.883...$
4
Round correctly
$O = 6.88$ cm (3 s.f.)
The side opposite 35Β° is 6.88 cm
Grade 6–7 Β· Cosine Rule & Area
In triangle $PQR$, $PQ = 8$ cm, $PR = 11$ cm and angle $P = 62Β°$. Find (a) the length $QR$, and (b) the area of the triangle.
1
Identify the formula β€” SAS, so cosine rule
We have two sides ($b = PQ = 8$, $c = PR = 11$) and the included angle ($A = P = 62Β°$). The side opposite $P$ is $QR = a$. Use $a^2 = b^2 + c^2 - 2bc\cos A$.
2
Substitute values
$QR^2 = 8^2 + 11^2 - 2(8)(11)\cos 62Β°$
$= 64 + 121 - 176\cos 62Β°$
$= 185 - 176 \times 0.4695...$
$= 185 - 82.63...$
$= 102.37...$
3
Square root
$QR = \sqrt{102.37...} = 10.118...$, so $QR = 10.1$ cm (3 s.f.)
4
Find the area using Β½ab sinC
Area $= \frac{1}{2} \times 8 \times 11 \times \sin 62Β° = 44 \times 0.8829... = 38.85...$
5
Round
Area $= 38.9$ cmΒ² (3 s.f.)
(a) $QR = 10.1$ cm     (b) Area $= 38.9$ cmΒ²
Grade 9 Β· 3D Trigonometry
A rectangular box (cuboid) has length 10 cm, width 6 cm and height 4 cm. Find the angle that the space diagonal makes with the base of the cuboid. Give your answer to 1 decimal place.
1
Find the base diagonal using Pythagoras
The base is a 10 Γ— 6 rectangle. The diagonal of the base $d$: $$d = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} = 2\sqrt{34}$$ Keep this exact β€” do not round yet.
2
Identify the right-angled triangle for the angle
The space diagonal runs from a bottom corner to the opposite top corner. The triangle formed has:
  • Base = base diagonal $= \sqrt{136}$ cm (horizontal, in the base)
  • Height = 4 cm (vertical, the height of the box)
  • Hypotenuse = space diagonal $= \sqrt{136 + 16} = \sqrt{152}$ cm
The angle $\theta$ is at the base corner, between the base diagonal and the space diagonal.
3
Apply trigonometry
$$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{\sqrt{136}}$$ $$\theta = \tan^{-1}\!\left(\frac{4}{\sqrt{136}}\right) = \tan^{-1}(0.3430...) = 18.94...Β°$$
4
Round to 1 decimal place
$\theta = 18.9Β°$ (1 d.p.)
The space diagonal makes an angle of 18.9Β° with the base.

❓ Exam Questions

Q1 1 mark

Write down the exact value of $\cos 60Β°$.

Answer: $\cos 60Β° = \dfrac{1}{2}$

Mark scheme: B1 for $\dfrac{1}{2}$ (or equivalent exact form). Decimal answers do not gain the mark as exact value is required.
Q2 2 marks

In a right-angled triangle, the adjacent side is 9.5 cm and the angle is 48Β°. Find the length of the hypotenuse. Give your answer to 3 significant figures.

Method: We have A and H, so use $\cos\theta = A/H$.
$\cos 48Β° = 9.5 / H$
$H = 9.5 / \cos 48Β° = 9.5 / 0.6691... = 14.19...$

Answer: $H = 14.2$ cm (3 s.f.)

Mark scheme: M1 for correct trig equation with cos; A1 for 14.2 (accept 14.19 to 14.20).
Q3 3 marks

Triangle $ABC$ has $AB = 7$ cm, $BC = 9$ cm and angle $ABC = 115Β°$. Calculate the area of triangle $ABC$.

Method: Sides $AB = 7$, $BC = 9$, included angle $B = 115Β°$.
Area $= \frac{1}{2} \times 7 \times 9 \times \sin 115Β°$
$= \frac{1}{2} \times 63 \times \sin 115Β°$
$= 31.5 \times 0.9063...$
$= 28.55...$

Answer: Area $= 28.5$ cmΒ² (3 s.f.) or $28.6$ cmΒ² depending on rounding

Mark scheme: M1 for correct formula with two sides and included angle; M1 for correct substitution; A1 for 28.5–28.6 cmΒ².
Q4 4 marks

In triangle $XYZ$, $XY = 14$ cm, $YZ = 11$ cm and angle $YXZ = 47Β°$. Use the sine rule to find angle $YZX$. Give your answer to 1 decimal place.

Method: Use $\dfrac{\sin Z}{XY} = \dfrac{\sin X}{YZ}$ (each angle over its opposite side).

Here $X = 47Β°$, side opposite $X$ is $YZ = 11$, side opposite $Z$ is $XY = 14$.
$\dfrac{\sin Z}{14} = \dfrac{\sin 47Β°}{11}$
$\sin Z = \dfrac{14 \times \sin 47Β°}{11} = \dfrac{14 \times 0.7314...}{11} = \dfrac{10.240...}{11} = 0.9309...$
$Z = \sin^{-1}(0.9309...) = 68.6Β°$

Check: since $\sin Z > 0$ and could be obtuse: $Z' = 180Β° - 68.6Β° = 111.4Β°$. Then $X + Z' = 47Β° + 111.4Β° = 158.4Β°$, leaving $Y = 21.6Β°$ β€” valid. Both solutions are geometrically possible unless additional context excludes one.

Answer: Angle $YZX = 68.6Β°$ (or $111.4Β°$ if obtuse solution stated)

Mark scheme: M1 correct sine rule set-up; M1 correct rearrangement; A1 for 68.6Β°; B1 for identifying the ambiguous case or confirming both solutions.
Q5 6 marks

A ship sails 15 km on a bearing of 040Β° from port $A$ to point $B$. It then sails 22 km on a bearing of 130Β° to reach port $C$. Calculate the direct distance from $A$ to $C$, and the bearing of $C$ from $A$. Give distances to 3 significant figures and bearings to the nearest degree.

Step 1 β€” Find angle ABC.
Bearing from $A$ to $B$ is 040Β°. At $B$, draw a North line. The back-bearing from $B$ to $A$ is 040Β° + 180Β° = 220Β°. Bearing from $B$ to $C$ is 130Β°. Interior angle at $B$ = 220Β° βˆ’ 130Β° = 90Β°. (The angle between $BA$ and $BC$ is 90Β°.)

Step 2 β€” Find AC using the cosine rule (SAS).
$AB = 15$, $BC = 22$, angle $B = 90Β°$.
$AC^2 = 15^2 + 22^2 - 2(15)(22)\cos 90Β° = 225 + 484 - 0 = 709$
(Since $\cos 90Β° = 0$, this reduces to Pythagoras.)
$AC = \sqrt{709} = 26.6$ km (3 s.f.)

Step 3 β€” Find angle BAC using the sine rule.
$\dfrac{\sin(\angle BAC)}{BC} = \dfrac{\sin B}{AC}$  β†’  $\sin(\angle BAC) = \dfrac{22 \times \sin 90Β°}{26.63...} = \dfrac{22}{26.63...} = 0.8261...$
$\angle BAC = 55.6Β°$

Step 4 β€” Find bearing of C from A.
Bearing of $B$ from $A$ is 040Β°. The direction from $A$ to $C$ is 040Β° + 55.6Β° = 095.6Β° β‰ˆ 096Β°.

Answer: $AC = 26.6$ km; Bearing of $C$ from $A$ = 096Β°

Mark scheme: B1 angle at B = 90Β°; M1 cosine rule or Pythagoras; A1 AC = 26.6 km; M1 sine rule for angle BAC; A1 55.6Β°; A1 bearing 096Β°. (6 marks)

⭐ Grade 9 Model Answers

Full Annotated Solution β€” Bearings Question (Q5)

✏️
WORKED EXAMPLE β€” What Grade 9 Looks Like

Question: A ship sails 15 km on a bearing of 040Β° from port $A$ to $B$, then 22 km on a bearing of 130Β° to port $C$. Find $AC$ and the bearing of $C$ from $A$.

[Mark-earning annotation 1 β€” Diagram with North arrows] βœ“
Drawing North arrows at $A$ and $B$ is the first step. A Grade 9 response always includes a clear diagram. At $B$, the North arrow points up. The back-bearing to $A$ is 040Β° + 180Β° = 220Β°, measured clockwise from North at $B$. The bearing from $B$ to $C$ is 130Β°. The angle $\angle ABC$ (interior) is found by subtracting: since the direction to $C$ (130Β°) is measured from North clockwise, and the direction to $A$ (220Β°) is also from North clockwise, the angle between them is 220Β° βˆ’ 130Β° = 90Β°.

[Mark-earning annotation 2 β€” Identifying the right formula] βœ“
With $AB = 15$, $BC = 22$, $\angle B = 90Β°$, this is a SAS scenario. A Grade 9 student recognises that $\cos 90Β° = 0$, so the cosine rule reduces to Pythagoras β€” demonstrating connections between topics. $AC^2 = 15^2 + 22^2 = 709$, so $AC = \sqrt{709} = 26.627...$ km. The unrounded value must be used for subsequent calculations.

[Mark-earning annotation 3 β€” Correct rounding at the end only] βœ“
$AC = 26.6$ km (3 s.f.). A Grade 9 student rounds only in the final step. Premature rounding is penalised at higher tier.

[Mark-earning annotation 4 β€” Finding angle BAC] βœ“
Apply the sine rule: $\dfrac{\sin(\angle BAC)}{22} = \dfrac{\sin 90Β°}{26.627...}$. Since $\sin 90Β° = 1$: $\sin(\angle BAC) = \dfrac{22}{26.627...} = 0.8262...$, giving $\angle BAC = 55.6Β°$. Explicitly stating the sine rule, showing the substitution, and then solving β€” this is what earns the method mark even if arithmetic is slightly off.

[Mark-earning annotation 5 β€” Computing the bearing correctly] βœ“
Bearing of $B$ from $A$ = 040Β°. Adding $\angle BAC = 55.6Β°$ (since $C$ is further clockwise than $B$ when viewed from $A$) gives bearing of $C$ from $A$ = 040Β° + 55.6Β° = 095.6Β° β‰ˆ 096Β°. A Grade 9 student shows this reasoning β€” not just the final answer β€” to secure the final mark.

🎯
EXAM TIP β€” What Distinguishes Grade 9 from Grade 7
  • Diagrams: Always draw, always label with North arrows for bearings problems.
  • Exact values: Use $\sqrt{709}$ not $26.6$ in intermediate calculations.
  • Justification: State which rule you are using and why, not just substituting numbers.
  • Ambiguous case: In SSA sine rule problems, check whether the obtuse solution is also valid.
  • Connection to other topics: Recognise when $\cos 90Β° = 0$ simplifies the cosine rule.

πŸ“‹ Revision Sheet

Key Definitions
HypotenuseLongest side; opposite the right angle
OppositeSide facing the angle in question
AdjacentSide next to the angle (not hypotenuse)
Included angleAngle directly between two known sides
Ambiguous caseTwo possible triangles from SSA data
BearingAngle measured clockwise from North, 3 digits
Essential Formulae

$\sin\theta = O/H \quad \cos\theta = A/H \quad \tan\theta = O/A$

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

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

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

Area $= \tfrac{1}{2}ab\sin C$

Space diagonal $= \sqrt{l^2+w^2+h^2}$

Memory Hooks
  • SOH-CAH-TOA β€” Some Old Hippos Can Always Hide Their Old Age
  • sin grows 0β†’1 from 0Β° to 90Β° (think of the sun rising)
  • 30-60-90 triangle sides: 1, √3, 2
  • 45-45-90 triangle sides: 1, 1, √2
  • Cosine rule β€” "Pythagoras with a correction term $-2bc\cos A$"
  • Area Β½ab sinC β€” C is the angle Caught between a and b
  • Ambiguous case β€” "Two triangles might hide in SSA"
Exam Tips
  • Always draw and label a diagram β€” even for non-right triangles
  • Use exact values (surds/fractions) unless decimals are required
  • Never round intermediate values β€” use calculator memory (ANS key)
  • Check: all angles in a triangle sum to 180Β°
  • In 3D problems, find the 2D cross-section triangle first
  • For bearings, draw North arrows and use alternate/co-interior angles
  • In the sine rule: put the unknown on the top of the fraction
  • State formulae before substituting β€” earns method marks

πŸ”„ Flashcards

Click each card to reveal the answer. 15 cards covering all key concepts.

βœ— Common Mistakes

βœ—
MISTAKE 1 β€” Labelling sides incorrectly
What students do wrong: Labelling O, A, H relative to the wrong angle, or confusing which angle is being used.
Why marks are lost: Substituting into sin/cos/tan with incorrect side labels gives wrong values β€” method marks are only awarded for a correct setup.
How to avoid it: Always circle the angle you are working with first, then label O (opposite), A (adjacent) and H (hypotenuse) relative to that angle before writing any formula.
βœ—
MISTAKE 2 β€” Using degrees instead of radians (or vice versa)
What students do wrong: Calculator is set to radians when degrees are needed (or vice versa), giving completely wrong trigonometric values.
Why marks are lost: All subsequent calculations are incorrect; no accuracy marks awarded even if method is correct.
How to avoid it: Check your calculator mode at the start of every trigonometry question. GCSE always uses degrees unless stated otherwise. The mode should read "DEG" or "D".
βœ—
MISTAKE 3 β€” Rounding too early in multi-step problems
What students do wrong: Rounding intermediate answers (e.g. rounding $\sqrt{136}$ to 11.7 before further calculation).
Why marks are lost: Accumulated rounding error causes the final answer to be outside the acceptable range; accuracy mark lost even with correct method.
How to avoid it: Keep full calculator display values throughout. Store intermediate results using the calculator's memory or ANS key. Only round at the final step.
βœ—
MISTAKE 4 β€” Using the sine rule when the cosine rule is needed
What students do wrong: Attempting the sine rule for SAS or SSS problems, where the sine rule cannot be applied without more information.
Why marks are lost: The method is fundamentally wrong; no method marks awarded even if arithmetic is correct.
How to avoid it: Identify the given information category first: SAS or SSS β†’ cosine rule; AAS, ASA, or SSA β†’ sine rule. Use the decision tree on this page as practice.
βœ—
MISTAKE 5 β€” Forgetting the ambiguous case of the sine rule
What students do wrong: When using the sine rule in SSA scenarios, finding only the acute angle and missing the obtuse solution.
Why marks are lost: In questions that ask you to "show all possible solutions" or where the obtuse angle is the intended answer, the mark cannot be earned without it.
How to avoid it: After finding $\sin B = k$ (where $k < 1$), always check whether $B' = 180Β° - B$ is also valid by verifying $A + B' < 180Β°$.
βœ—
MISTAKE 6 β€” Using non-included angle in the area formula
What students do wrong: Writing Area $= \frac{1}{2} \times AB \times BC \times \sin A$ (using angle $A$, which is opposite $BC$, not between $AB$ and $BC$).
Why marks are lost: The formula gives an incorrect area; method mark lost if the set-up is clearly wrong.
How to avoid it: In Area $= \frac{1}{2}ab\sin C$, the angle $C$ must be the angle at the vertex between the two sides $a$ and $b$. Identify the vertex where the two known sides meet β€” that angle is $C$.

βœ… Final Checklist

Click each item to mark it as complete. Progress is saved automatically.

  • I can label O, A and H relative to any angle in a right-angled triangle
  • I can apply sin, cos and tan to find a missing side in a right-angled triangle
  • I can apply sin⁻¹, cos⁻¹ and tan⁻¹ to find a missing angle in a right-angled triangle
  • I know the exact values of sin, cos and tan for 0Β°, 30Β°, 45Β°, 60Β° and 90Β°
  • I can derive exact values from the 30-60-90 and 45-45-90 special triangles
  • I can apply the sine rule to find a missing side (AAS or ASA)
  • I can apply the sine rule to find a missing angle and identify the ambiguous case
  • I can apply the cosine rule to find a missing side (SAS)
  • I can apply the cosine rule to find a missing angle (SSS)
  • I can use Area = Β½ab sinC to find the area of any triangle
  • I can identify and extract a right-angled triangle from a 3D shape for trig problems
  • I can find the space diagonal of a cuboid using Pythagoras in two stages
  • I can convert between bearings and interior angles of a triangle
  • I can decide which formula (SOH-CAH-TOA, sine rule, cosine rule) to use for any given triangle
  • I know to check my calculator is in DEG mode before every trigonometry question
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