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 a surveyor measuring a hillside, a pilot calculating flight paths, or a GCSE/A-Level question worth 5–8 marks. Miss this, and you’re leaving easy marks on the table."

What You Need To Know First

1. Labelling triangles: Sides a, b, c opposite angles A, B, C (standard notation).
2. Pythagoras’ Theorem and trigonometry in right-angled triangles (SOHCAHTOA).
3. Rearranging equations (e.g., solving for a variable in a = b × sin(C)).

Key Vocabulary

Formulas To Know

1. The Sine Rule

Formula: [ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} ] Variables: - a, b, c = lengths of sides opposite angles A, B, C. - A, B, C = angles in degrees or radians.

When to use: - Two angles + one side (find a missing side). - Two sides + a non-included angle (find a missing angle).

MEMORISE THIS (not always given on 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} ] Variables: - a = side opposite angle A. - b, c = other two sides.

When to use: - Three sides (find an angle). - Two sides + the included angle (find the third side).

MEMORISE THIS (sometimes given, but not always).

3. Area of a Triangle (Using Sine)

Formula: [ \text{Area} = \frac{1}{2} ab \sin C ] Variables: - a, b = two sides. - C = the included angle between them.

When to use: - When you know two sides and the included angle (no height needed).

GIVEN ON EXAM SHEET (but memorise it anyway—it saves time).

Step-by-Step Method

When to Use Which Rule?

Step-by-Step: Sine Rule

Problem: Find side x in △ABC where A = 40°, B = 60°, and a = 5 cm.

1. Label the triangle (side a opposite angle A, etc.).
2. Check what you know: Two angles (A, B) and one side (a).
3. Find the missing angle (if needed):
   [
   C = 180° - A - B = 180° - 40° - 60° = 80°
   ]
4. Write the Sine Rule (pick two ratios):
   [
   \frac{a}{\sin A} = \frac{x}{\sin B}
   ]
5. Plug in known values:
   [
   \frac{5}{\sin 40°} = \frac{x}{\sin 60°}
   ]
6. Solve for x:
   [
   x = \frac{5 \times \sin 60°}{\sin 40°}
   ]
7. Calculate (use a calculator):
   [
   x ≈ \frac{5 \times 0.8660}{0.6428} ≈ 6.74 \text{ cm}
   ]
8. Round to 3 significant figures (if required): 6.74 cm.

Why this works: The Sine Rule links sides and angles proportionally—like a "trig version" of similar triangles.

Step-by-Step: Cosine Rule

Problem: Find angle A in △ABC where a = 7 cm, b = 5 cm, c = 4 cm.

1. Label the triangle (side a opposite angle A, etc.).
2. Check what you know: Three sides (a, b, c).
3. Write the Cosine Rule for angles:
   [
   \cos A = \frac{b^2 + c^2 - a^2}{2bc}
   ]
4. Plug in values:
   [
   \cos A = \frac{5^2 + 4^2 - 7^2}{2 \times 5 \times 4} = \frac{25 + 16 - 49}{40} = \frac{-8}{40} = -0.2
   ]
5. Find angle A (use inverse cosine):
   [
   A = \cos^{-1}(-0.2) ≈ 101.54°
   ]
6. Round to 1 decimal place: 101.5°.

Why this works: The Cosine Rule is like Pythagoras’ Theorem with an extra term to account for non-right angles.

Worked Examples

Example 1 – Basic (Sine Rule)

Question: In △PQR, P = 35°, Q = 85°, and p = 12 cm. Find side q.

Solution: 1. Find angle R:
[
R = 180° - 35° - 85° = 60°
] 2. Write Sine Rule:
[
\frac{p}{\sin P} = \frac{q}{\sin Q}
] 3. Plug in values:
[
\frac{12}{\sin 35°} = \frac{q}{\sin 85°}
] 4. Solve for q:
[
q = \frac{12 \times \sin 85°}{\sin 35°} ≈ \frac{12 \times 0.9962}{0.5736} ≈ 20.8 \text{ cm}
]

What we did and why: - Used the Sine Rule because we had two angles and one side. - Found the missing angle first to complete the ratio.

Example 2 – Medium (Cosine Rule + Ambiguous Case)

Question: In △XYZ, x = 8 cm, y = 6 cm, and X = 50°. Find angle Y.

Solution: 1. Write Sine Rule:
[
\frac{x}{\sin X} = \frac{y}{\sin Y}
] 2. Plug in values:
[
\frac{8}{\sin 50°} = \frac{6}{\sin Y}
] 3. Solve for sin Y:
[
\sin Y = \frac{6 \times \sin 50°}{8} ≈ 0.5745
] 4. Find Y (two possible angles!):
[
Y ≈ \sin^{-1}(0.5745) ≈ 35.1° \quad \text{or} \quad 180° - 35.1° = 144.9°
] 5. Check validity:
- If Y = 144.9°, then X + Y = 50° + 144.9° = 194.9° > 180° → invalid.
- Only Y = 35.1° is possible.

What we did and why: - Used the Sine Rule but checked for the ambiguous case (two possible angles). - Eliminated the invalid angle by ensuring the triangle’s angles sum to 180°.

Example 3 – Exam-Style (Area + Cosine Rule)

Question: A triangle has sides a = 9 cm, b = 7 cm, and c = 5 cm. a) Find angle C. b) Find the area of the triangle.

Solution (a): 1. Write Cosine Rule for angle C:
[
\cos C = \frac{a^2 + b^2 - c^2}{2ab}
] 2. Plug in values:
[
\cos C = \frac{9^2 + 7^2 - 5^2}{2 \times 9 \times 7} = \frac{81 + 49 - 25}{126} = \frac{105}{126} ≈ 0.8333
] 3. Find C:
[
C = \cos^{-1}(0.8333) ≈ 33.6°
]

Solution (b): 1. Use area formula (two sides + included angle):
[
\text{Area} = \frac{1}{2} ab \sin C
] 2. Plug in values:
[
\text{Area} = \frac{1}{2} \times 9 \times 7 \times \sin 33.6° ≈ \frac{1}{2} \times 63 \times 0.553 ≈ 17.4 \text{ cm}^2
]

What we did and why: - Used the Cosine Rule first to find an angle, then the area formula with that angle. - No height needed—the formula does the work for us.

Common Mistakes

Exam Traps

1-Minute Recap

"Here’s the night-before cheat sheet: 1. Sine Rule: Use when you have two angles + one side or two sides + a non-included angle. Write the ratio, plug in numbers, and solve. 2. Cosine Rule: Use for three sides or two sides + the included angle. Rearrange to find sides or angles. 3. Area formula: ½ ab sin C – no height needed! 4. Ambiguous case: If sin(θ) gives two angles, check if both fit in the triangle. 5. Label carefully: Side a is always opposite angle A.

Now go smash those 6-mark questions!