# Trigonometry and Pythagoras' theorem

> Edexcel International GCSE Mathematics A · 4MA1
> Source: https://www.owlsprep.com/study/edexcel-igcse-math-a-s4-trigonometry-and-pythagoras-theorem/

This guide covers all Foundation and Higher content for Edexcel IGCSE Maths A Trigonometry and Pythagoras' theorem, including 2D/3D problems, trig ratios, sine/cosine rules, bearings, and triangle area calculations for exam success.

**Prerequisites:** [Basic properties of triangles and angles](https://www.owlsprep.com/study/edexcel-igcse-math-a-s4-basic-geometry/); [Calculator use for square roots and trig functions](https://www.owlsprep.com/study/edexcel-igcse-math-a-s1-calculator-skills/)

## Learning objectives

- Use Pythagoras' theorem to solve 2D (Foundation) and 3D (Higher) right-triangle problems
- Apply SOH-CAH-TOA for acute angles in right-angled triangles (Foundation) and obtuse angles (Higher)
- Solve 2D trig problems including bearings (Foundation) and angles of elevation/depression (Higher)
- Use sine rule, cosine rule, and (1/2)ab sin C area formula for non-right triangles (Higher)
- Solve 3D trig problems including line-plane angles (Higher)

## Pythagoras' Theorem for 2D and 3D Triangles

**Pythagoras' Theorem** — For a right-angled triangle with hypotenuse (longest side, opposite right angle) $c$, and shorter sides $a$ and $b$, the sum of squares of the shorter sides equals the square of the hypotenuse.

*Notation:* a^2 + b^2 = c^2

*Example:* Right triangle with sides 3cm, 4cm: hypotenuse = 5cm, since $3^2 +4^2 =5^2$.

Foundation tier only covers 2D applications of Pythagoras' theorem, while Higher tier extends this to 3D shapes (cuboids, pyramids, prisms) by identifying right triangles inside the 3D shape.

**Worked example:** A cuboid has length 6cm, width 4cm, height 3cm. Find the length of the space diagonal of the cuboid (Higher only).

1. First find the diagonal of the base rectangle using 2D Pythagoras: base diagonal $d_1^2 = 6^2 +4^2 = 36 +16 =52$.
2. Form a right triangle with $d_1$, height 3cm, and space diagonal $d$: $d^2 = d_1^2 + 3^2 =52 +9 =61$.
3. Calculate $d = \sqrt{61} \approx7.81$cm (3 significant figures).

> **tip**
>
> For 3D Pythagoras for cuboids, you can use the direct formula $d = \sqrt{l^2 +w^2 +h^2}$, but always show your working for full marks.

> **Exam tip:** Always label the hypotenuse clearly to avoid mixing up sides when applying Pythagoras, and check your final value is longer than the other two sides.

*Calculator:* allowed

## Right-Angled Trigonometry (SOH-CAH-TOA)

**SOH-CAH-TOA** — Mnemonic for the three trigonometric ratios for acute angles in right-angled triangles: SOH: $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$; CAH: $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$; TOA: $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$.

> **Trig Ratio Memory Hook**
>
> SOH-CAH-TOA

Foundation tier uses these ratios only for acute angles in 2D right triangles, while Higher tier extends use to obtuse angles (note: $\sin(\text{obtuse}) = \sin(180-\theta)$, $\cos(\text{obtuse})$ is negative).

**Worked example:** A right triangle has hypotenuse 12cm, and an acute angle of 35°. Find the length of the side opposite the 35° angle.

1. Label sides: hypotenuse =12cm, opposite side =$x$, angle $\theta=35^\circ$.
2. We have hypotenuse and need opposite, so use sine (SOH): $\sin 35^\circ = \frac{x}{12}$.
3. Rearrange: $x =12 \times \sin 35^\circ \approx12 \times 0.5736 \approx6.88$cm (3 sf).

> **Exam tip:** For Higher tier, if you calculate an obtuse angle using inverse sine, remember to check if 180° minus your result is the correct answer for the triangle.

*Calculator:* allowed

## 2D Trigonometry Problem Solving

Foundation tier problems include bearings (three-digit angles measured clockwise from north), while Higher tier adds angles of elevation (angle from horizontal up to an object) and depression (angle from horizontal down to an object, equal to elevation due to alternate angles).

**Worked example:** A ship sails 15km east from port A, then 8km north to port B. Find the bearing of B from A (Foundation tier).

1. Draw a right triangle with east side 15km, north side 8km, right angle at the turning point.
2. Find the angle between the north line and AB: $\tan \theta = \frac{15}{8} = 1.875$.
3. Calculate $\theta = \arctan(1.875) \approx61.9^\circ$, so bearing = 062° (three-digit format).

> **info**
>
> Bearings are always written as three digits, so 62° becomes 062° for full marks.

*Calculator:* allowed

## Sine Rule, Cosine Rule and Triangle Area Formula (Higher Only)

For non-right angled triangles, use the sine rule, cosine rule, and area formula provided on your exam formula sheet. Use the sine rule when you have a matching side-angle pair, and the cosine rule when you have three sides or two sides and an included angle.

**Worked example:** A triangle has sides $a=7$cm, $b=9$cm, and included angle $C=40^\circ$. Find the area of the triangle, and the length of side $c$.

1. Calculate area: $\frac{1}{2}ab \sin C = 0.5 \times7\times9 \times \sin40^\circ \approx20.2$ cm² (3 sf).
2. Use cosine rule to find $c$: $c^2 = a^2 +b^2 - 2ab \cos C = 7^2 +9^2 - 2\times7\times9\times\cos40^\circ$.
3. Simplify: $c^2 = 49 +81 - 126 \times 0.7660 \approx33.5$, so $c \approx5.79$ cm (3 sf).

> **Exam tip:** The ambiguous case of the sine rule only occurs when you are finding an angle opposite a shorter side, so always check both possible acute and obtuse solutions.

*Calculator:* allowed

## 3D Trigonometry (Higher Only)

Higher tier requires you to find the angle between a line and a plane. This is the smallest angle between the line and its projection onto the plane, forming a right triangle with the line, its projection, and the perpendicular height from the end of the line to the plane.

**Worked example:** A square-based pyramid has base side length 10cm, and vertical height 12cm. Find the angle between one of the slant edges and the base plane.

1. The projection of the slant edge onto the base is half the diagonal of the square base: half-diagonal = $5\sqrt{2} \approx7.071$cm.
2. The right triangle has opposite side = height 12cm, adjacent = $5\sqrt{2}$ cm, angle $\theta$ between slant edge and base.
3. $\tan \theta = \frac{12}{5\sqrt{2}} \approx1.697$, so $\theta \approx59.5^\circ$ (1 dp).

> **Exam tip:** Always identify the right triangle explicitly in 3D problems, labelling all three sides clearly to avoid mixing up trig ratios.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Mixing up the hypotenuse with a shorter side when applying Pythagoras' theorem.
  - Why it fails: The hypotenuse is always the longest side, so it must be the $c$ term in $a^2 + b^2 = c^2$.
  - Correct: Label the hypotenuse (opposite the right angle) first, then plug values into the formula, checking that $c$ is the largest value.
- **Wrong:** Using calculator in radians mode for trig calculations.
  - Why it fails: All Edexcel IGCSE Maths A angles are in degrees, so radian mode will give incorrect values for sin, cos, tan.
  - Correct: Check your calculator shows "D" or "DEG" on the display before starting any trig question.
- **Wrong:** Rounding intermediate calculation values too early.
  - Why it fails: Early rounding leads to inaccuracies in the final answer, which can cost marks.
  - Correct: Keep full calculator precision for all intermediate steps, only round the final answer to the required number of significant figures (usually 3 sf) or decimal places.
- **Wrong:** Forgetting to write bearings as three-digit numbers.
  - Why it fails: Bearings are defined as three-digit angles measured clockwise from north, so missing leading zeros will lose marks.
  - Correct: Add leading zeros to bearings less than 100°, e.g. 62° becomes 062°, 8° becomes 008°.
- **Wrong:** Only considering the acute angle solution when using the sine rule for obtuse angles.
  - Why it fails: $\sin \theta = \sin(180^\circ-\theta)$, so there are two possible angles for any positive sine value between 0 and 1.
  - Correct: Check if the obtuse solution (180° minus the acute result) sums to less than 180° with the other known angles in the triangle, and use the appropriate value.
- **Wrong:** Using non-included angles for the $\frac{1}{2}ab \sin C$ area formula.
  - Why it fails: The formula only works when angle $C$ is the included angle between sides $a$ and $b$.
  - Correct: Ensure the angle you use is between the two known sides, or rearrange the triangle to use an included angle if needed.

## Cheatsheet

| Concept | Formula/Rule | Tier | Recall/Given |
| --- | --- | --- | --- |
| Pythagoras' Theorem (2D) | $a^2 + b^2 = c^2$, $c$ = hypotenuse | Foundation | Recall |
| SOH-CAH-TOA Ratios | $\sin\theta=O/H$, $\cos\theta=A/H$, $\tan\theta=O/A$ | Foundation | Recall |
| Bearings | 3-digit, clockwise from north | Foundation | Recall |
| Pythagoras' Theorem (3D cuboid) | $d = \sqrt{l^2 + w^2 + h^2}$ | Higher | Recall/Derive |
| Sine Rule | $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ | Higher | Given |
| Cosine Rule | $a^2 = b^2 + c^2 - 2bc \cos A$ | Higher | Given |
| Triangle Area (non-right) | $\frac{1}{2}ab \sin C$ (included angle $C$) | Higher | Given |
| Angle of elevation/depression | Equal due to alternate angles | Higher | Recall |

## What's next

Now that you have mastered trigonometry and Pythagoras' theorem for Edexcel IGCSE Maths A, you are ready to apply these skills to more complex geometry problems, including circle theorems and vector calculations, which often combine with trigonometry in higher-mark exam questions. Be sure to practice as many past paper questions as possible for both Foundation and Higher tiers, focusing on showing all your working clearly to maximise marks, and always checking your answers for reasonableness (e.g., the hypotenuse should always be the longest side of a right triangle). Pay special attention to ambiguous sine rule cases and 3D trigonometry problems, which are common discriminators for higher grades in the Higher tier paper.

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