GCSE Maths Geometry and Measures - How to Solve: The Sine and Cosine Rules (Non-Right-Angled Triangles)

How to Solve: The Sine and Cosine Rules (Non-Right-Angled Triangles)

Introduction

"Master the Sine and Cosine Rules, and you’ll solve any triangle—whether it’s calculating the height of a tree in Physics, the bond angle in Chemistry, or the distance between two points in Biology. This topic appears in 10-15% of GCSE/A-Level Maths papers, and missing it could cost you 12-18 marks—enough to drop a grade. Let’s make sure you never lose those marks again."

WHAT YOU NEED TO KNOW FIRST

Before starting, you must understand: 1. Basic trigonometry (SOHCAHTOA) – You should know how to use sine, cosine, and tangent in right-angled triangles. 2. Labelling triangles correctly – Sides are labelled a, b, c opposite angles A, B, C (standard notation). 3. Rearranging formulas – You’ll need to solve for unknowns (angles or sides).

If you’re shaky on any of these, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

• Non-right-angled triangle: A triangle with no 90° angle.
• Opposite side: The side across from the angle you’re working with.
• Included angle: The angle between two sides.

Formulas

1. The Sine Rule

Formula: [ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} ] What it means: - Use when you know two angles and one side (AAS or ASA) OR two sides and a non-included angle (SSA). - MEMORISE THIS – It’s not given on most exam sheets.

2. The Cosine Rule

Formula (for sides): [ a^2 = b^2 + c^2 - 2bc \cos A ] Formula (for angles): [ \cos A = \frac{b^2 + c^2 - a^2}{2bc} ] What it means: - Use when you know three sides (SSS) OR two sides and the included angle (SAS). - MEMORISE THIS – The side version is not always given; the angle version is rarely given.

STEP-BY-STEP METHOD

When to Use Which Rule?

Step-by-Step: Using the Sine Rule

Problem: Find side a in a triangle where: - Angle A = 40° - Angle B = 60° - Side b = 7 cm

Steps: 1. Label the triangle – Write down all known values (angles and sides). 2. Check if you can use the Sine Rule – Do you have two angles and one side? Yes. 3. Write the Sine Rule formula:
[
\frac{a}{\sin A} = \frac{b}{\sin B}
] 4. Plug in known values:
[
\frac{a}{\sin 40°} = \frac{7}{\sin 60°}
] 5. Rearrange to solve for a:
[
a = \frac{7 \times \sin 40°}{\sin 60°}
] 6. Calculate (use a calculator):
[
a ≈ \frac{7 \times 0.6428}{0.8660} ≈ 5.2 \text{ cm}
] 7. Round to 1 decimal place (unless the question specifies otherwise).

What we did and why: - We used the Sine Rule because we had two angles and one side. - We rearranged the formula to isolate the unknown side a. - Always check your calculator is in degree mode!

Step-by-Step: Using the Cosine Rule (Finding a Side)

Problem: Find side a in a triangle where: - Side b = 5 cm - Side c = 8 cm - Angle A = 50°

Steps: 1. Label the triangle – Write down all known values. 2. Check if you can use the Cosine Rule – Do you have two sides and the included angle (SAS)? Yes. 3. Write the Cosine Rule formula for sides:
[
a^2 = b^2 + c^2 - 2bc \cos A
] 4. Plug in known values:
[
a^2 = 5^2 + 8^2 - 2 \times 5 \times 8 \times \cos 50°
] 5. Calculate step by step:
- (5^2 = 25)
- (8^2 = 64)
- (2 \times 5 \times 8 = 80)
- (\cos 50° ≈ 0.6428)
- (80 \times 0.6428 ≈ 51.424) 6. Substitute back:
[
a^2 = 25 + 64 - 51.424 = 37.576
] 7. Take the square root:
[
a ≈ \sqrt{37.576} ≈ 6.1 \text{ cm}
] 8. Round to 1 decimal place.

What we did and why: - We used the Cosine Rule because we had two sides and the included angle. - We calculated step by step to avoid mistakes. - Always check units (cm, m, etc.) and round at the end.

Step-by-Step: Using the Cosine Rule (Finding an Angle)

Problem: Find angle A in a triangle where: - Side a = 6 cm - Side b = 5 cm - Side c = 8 cm

Steps: 1. Label the triangle – Write down all known sides. 2. Check if you can use the Cosine Rule – Do you have three sides (SSS)? Yes. 3. Write the Cosine Rule formula for angles:
[
\cos A = \frac{b^2 + c^2 - a^2}{2bc}
] 4. Plug in known values:
[
\cos A = \frac{5^2 + 8^2 - 6^2}{2 \times 5 \times 8}
] 5. Calculate step by step:
- (5^2 = 25)
- (8^2 = 64)
- (6^2 = 36)
- (2 \times 5 \times 8 = 80) 6. Substitute back:
[
\cos A = \frac{25 + 64 - 36}{80} = \frac{53}{80} = 0.6625
] 7. Find angle A using inverse cosine:
[
A = \cos^{-1}(0.6625) ≈ 48.5°
] 8. Round to 1 decimal place.

What we did and why: - We used the Cosine Rule because we had three sides. - We rearranged the formula to solve for the angle. - Always use inverse cosine (cos⁻¹) to find the angle.

WORKED EXAMPLES

Example 1 – Basic (Sine Rule)

Problem: In triangle ABC, angle A = 35°, angle B = 70°, and side b = 12 cm. Find side a.

Solution: 1. Label the triangle:
- A = 35°, B = 70°, b = 12 cm 2. Check rule: Two angles + one side → Sine Rule. 3. Write formula:
[
\frac{a}{\sin 35°} = \frac{12}{\sin 70°}
] 4. Rearrange:
[
a = \frac{12 \times \sin 35°}{\sin 70°}
] 5. Calculate:
[
a ≈ \frac{12 \times 0.5736}{0.9397} ≈ 7.3 \text{ cm}
]

What we did and why: - Used the Sine Rule because we had two angles and one side. - Rearranged to solve for a and calculated carefully.

Example 2 – Medium (Cosine Rule – Finding a Side)

Problem: In triangle PQR, PQ = 9 cm, PR = 7 cm, and angle P = 45°. Find QR.

Solution: 1. Label the triangle:
- PQ = 9 cm, PR = 7 cm, angle P = 45° 2. Check rule: Two sides + included angle → Cosine Rule. 3. Write formula:
[
QR^2 = 9^2 + 7^2 - 2 \times 9 \times 7 \times \cos 45°
] 4. Calculate:
- (9^2 = 81)
- (7^2 = 49)
- (2 \times 9 \times 7 = 126)
- (\cos 45° ≈ 0.7071)
- (126 \times 0.7071 ≈ 89.09) 5. Substitute back:
[
QR^2 = 81 + 49 - 89.09 = 40.91
] 6. Square root:
[
QR ≈ \sqrt{40.91} ≈ 6.4 \text{ cm}
]

What we did and why: - Used the Cosine Rule because we had two sides and the included angle. - Calculated step by step to avoid errors.

Example 3 – Exam-Style (Mixed Rules)

Problem: A ship sails 10 km on a bearing of 040° from port A to port B. It then changes course to 120° and sails 8 km to port C. Find the direct distance from port A to port C.

Solution: 1. Draw the triangle:
- AB = 10 km, BC = 8 km
- Angle at B = 120° - 40° = 80° (bearings rule) 2. Check rule: Two sides + included angle → Cosine Rule. 3. Write formula:
[
AC^2 = 10^2 + 8^2 - 2 \times 10 \times 8 \times \cos 80°
] 4. Calculate:
- (10^2 = 100)
- (8^2 = 64)
- (2 \times 10 \times 8 = 160)
- (\cos 80° ≈ 0.1736)
- (160 \times 0.1736 ≈ 27.78) 5. Substitute back:
[
AC^2 = 100 + 64 - 27.78 = 136.22
] 6. Square root:
[
AC ≈ \sqrt{136.22} ≈ 11.7 \text{ km}
]

What we did and why: - Bearings gave us the angle between the two sides. - Used the Cosine Rule because we had two sides and the included angle. - Real-world application (Physics/Navigation).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

"Here’s what you need to remember the night before your exam: 1. Sine Rule: Use when you have two angles + one side or two sides + a non-included angle. Formula: (\frac{a}{\sin A} = \frac{b}{\sin B}). 2. Cosine Rule: Use when you have three sides or two sides + the included angle. Formula: (a^2 = b^2 + c^2 - 2bc \cos A). 3. Label your triangle properly – sides opposite angles, angles opposite sides. 4. Check your calculator mode – always in degrees for GCSE/A-Level. 5. Round at the end – don’t round intermediate steps. 6. Practice one problem of each type – Sine Rule (AAS), Cosine Rule (SAS), and Cosine Rule (SSS).

You’ve got this. Now go ace that exam!"