Trigonometry in two dimensions Revision Notes for AQA GCSE Further Maths

Trigonometry in two dimensions

Introduction to trigonometric ratios

Trigonometry helps us work with triangles, particularly right-angled triangles. The three main trigonometric functions - sine (sin), cosine (cos), and tangent (tan) - create relationships between the angles and sides of right triangles.

In any right triangle, we can identify three key parts relative to a given angle θ (theta):

Hypotenuse: The longest side, opposite the right angle
Opposite: The side directly across from the angle θ
Adjacent: The side next to angle θ (but not the hypotenuse)

The fundamental trigonometric ratios are defined as:

sin θ = opposite/hypotenuse
cos θ = adjacent/hypotenuse
tan θ = opposite/adjacent

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A helpful way to remember these ratios is using the mnemonic SOHCAHTOA:

Sine = Opposite over Hypotenuse
Cosine = Adjacent over Hypotenuse
Tangent = Opposite over Adjacent

Finding missing sides

When you know an angle and one side of a right triangle, you can use trigonometric ratios to find the other sides. The key is choosing the correct ratio based on what information you have and what you need to find.

Example: Finding the opposite side

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Worked Example: Finding the Opposite Side

To find side length 'a' when you know the hypotenuse is 12 cm and the angle is 35°:

Since we want the opposite side and we know the hypotenuse, we use sine: $\sin 35° = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{12}$

Rearranging: $a = 12 \times \sin 35°$ Therefore: $a = 6.9$ cm (to 1 decimal place)

Example: Finding the hypotenuse

When you know the adjacent side and need the hypotenuse, use cosine.

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Worked Example: Finding the Hypotenuse

If the adjacent side is 4.2 m and the angle is 18°:

$\cos 18° = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4.2}{\text{hypotenuse}}$

Rearranging: $\text{hypotenuse} = \frac{4.2}{\cos 18°}$ Therefore: $\text{hypotenuse} = 4.42$ m (to 3 significant figures)

Finding missing angles

When you know two sides of a right triangle, you can find the angles using inverse trigonometric functions ( $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ ).

Example: Using tangent to find an angle

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Worked Example: Finding an Angle Using Tangent

In a triangle where the opposite side is 5 cm and the adjacent side is 7 cm:

$\tan θ = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{7}$

To find θ: $θ = \tan^{-1}\left(\frac{5}{7}\right) = 35.5°$ (to 1 decimal place)

Real-world applications

Trigonometry is particularly useful for solving problems involving angles of elevation and depression. These angles are always measured from the horizontal line.

Angles of elevation and depression

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Angle of elevation: Measured upward from horizontal
Angle of depression: Measured downward from horizontal

Example: Bird flight problem

A bird flies from the top of a 15 m tree at an angle of depression of 27° to catch a worm on the ground.

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Worked Example: Bird Flight Problem

To find how far the bird flies (hypotenuse h):

The angle with the horizontal is $90° - 27° = 63°$
$\cos 63° = \frac{15}{h}$
$h = \frac{15}{\cos 63°} = 33.0$ m

To find the horizontal distance (x): Using Pythagoras' theorem: $x^2 = h^2 - 15^2$ $x = 29.4$ m from the bottom of the tree

Special angles: 30°, 45°, and 60°

These angles have exact trigonometric values that you should memorise for non-calculator questions.

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Essential Values to Memorise:

For a 45° angle (from an isosceles right triangle):

$\sin 45° = \frac{1}{\sqrt{2}}$
$\cos 45° = \frac{1}{\sqrt{2}}$
$\tan 45° = 1$

For 30° and 60° angles (from an equilateral triangle split in half):

$\sin 30° = \frac{1}{2}$ , $\cos 30° = \frac{\sqrt{3}}{2}$ , $\tan 30° = \frac{1}{\sqrt{3}}$
$\sin 60° = \frac{\sqrt{3}}{2}$ , $\cos 60° = \frac{1}{2}$ , $\tan 60° = \sqrt{3}$

Working with exact values

When solving problems with special angles, always give answers in exact form using surds rather than decimal approximations.

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Worked Example: Using Exact Values

For a 30° angle in a right triangle:

$\cos 30° = \frac{y}{8 + 6\sqrt{3}}$

Since $\cos 30° = \frac{\sqrt{3}}{2}$ : $\frac{\sqrt{3}}{2} = \frac{y}{8 + 6\sqrt{3}}$

Solving: $y = \frac{\sqrt{3}}{2} \times (8 + 6\sqrt{3}) = 4\sqrt{3} + 9$

Problem-solving strategies

Step-by-step approach

A systematic approach is essential for success in trigonometry problems:

Draw a clear diagram labelling all known information
Identify what you need to find and mark it clearly
Choose the appropriate trigonometric ratio based on the known and unknown sides
Set up the equation and solve algebraically
Check your answer makes sense in the context

Common exam tips

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Always check which sides are opposite, adjacent, and hypotenuse relative to the angle you're using
For angles of depression, remember they create alternate angles with angles of elevation
When working with exact values, rationalise denominators where necessary
Round final answers appropriately according to the question's requirements
Store full calculator values during multi-step calculations to avoid rounding errors

Avoiding common mistakes

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Common Pitfalls to Avoid:

Don't confuse opposite and adjacent sides - always identify them relative to your chosen angle
Remember that angles of depression are measured downward from horizontal, not from vertical
When using inverse trigonometric functions, ensure your calculator is in the correct mode (degrees or radians)
For exact value questions, don't use a calculator - work with the known exact ratios

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Key Points to Remember:

SOHCAHTOA helps you remember the basic trigonometric ratios: $\sin = \frac{O}{H}$ , $\cos = \frac{A}{H}$ , $\tan = \frac{O}{A}$
Special angle values must be memorised: $\sin 30° = \frac{1}{2}$ , $\cos 45° = \frac{1}{\sqrt{2}}$ , $\tan 60° = \sqrt{3}$
Always draw diagrams clearly labelling the sides as opposite, adjacent, and hypotenuse relative to your chosen angle
Angles of elevation look up from horizontal, while angles of depression look down from horizontal
Store full calculator values during multi-step problems to maintain accuracy in your final answer

Trigonometry in two dimensions (AQA GCSE Further Maths): Revision Notes