Sine & Cosine Rules
The sine and cosine rules unlock triangle problems that basic right-triangle trigonometry cannot solve. These laws help students find missing sides and angles in any triangle, making them essential tools for CCSS.HSG.SRT geometry standards.
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Why it matters
Engineering students use these rules to calculate structural loads in non-right triangles, while surveyors apply them to measure distances across irregular terrain. In navigation, pilots use the cosine rule to determine flight paths between 3 airports forming a triangle. Architecture students calculate roof angles and support beam lengths using both rules. A surveyor measuring a triangular plot with sides 150m, 200m, and an included angle of 75° needs the cosine rule to find the third side (approximately 235m). GPS systems rely on triangulation principles that mirror these fundamental laws, processing millions of triangle calculations daily to pinpoint locations within 3 meters of accuracy.
How to solve sine & cosine rules
Sine & Cosine Rules
- Law of sines: a/sin(A) = b/sin(B) = c/sin(C). Use for AAS or SSA.
- Law of cosines: c² = a² + b² − 2ab·cos(C). Use for SAS (find third side).
- Rearranged: cos(C) = (a² + b² − c²)/(2ab). Use for SSS (find an angle).
- Each side is paired with the sine of the angle opposite it.
Example: a=5, b=7, C=60° → c² = 25 + 49 − 70·(12) = 39, so c ≈ 6.24.
Worked examples
You are given two sides and the included angle of a triangle. Which rule applies, and what is its formula?
Answer: Law of cosines: c² = a² + b² − 2ab·cos(C)
- Recognise the SAS configuration → Scenario: SAS — AAS / SSA → sine rule. SAS / SSS → cosine rule.
- Write the formula → c² = a² + b² − 2ab·cos(C) — Use the law of cosines when this configuration is given.
In a triangle, side a = 10, angle A = 45°, angle B = 60°. Find side b.
Answer: b ≈ 12.25
- Identify the rule → AAS → law of sines — With two angles and a non-included side (AAS), the law of sines applies.
- Write the formula with given values → 10/sin(45°) = b/sin(60°) — Pair each side with the sine of its opposite angle.
- Solve for b → b = 10 · sin(60°) / sin(45°) = 10 · 0.866 / 0.7071 — Multiply both sides by sin(B) to isolate b.
- Approximate to 2 decimals → b ≈ 12.25 — Evaluate numerically to the requested precision.
In a triangle, side a = 8, side b = 5, and the included angle C = 60°. Find side c.
Answer: c ≈ 7.0
- Identify the rule → SAS → law of cosines — Two sides and the included angle → use the law of cosines.
- Write the formula with given values → c² = 8² + 5² − 2·8·5·cos(60°) — c² = a² + b² − 2ab·cos(C).
- Solve algebraically → c² = 64 + 25 − 80·0.5 = 49.0 — Compute each term, then combine.
- Take square root and round → c = √49.0 ≈ 7.0 — Side lengths are positive; round to 2 decimals.
Common mistakes
- Students confuse when to use each rule, applying sine rule to SAS problems like finding the third side when given a=6, b=8, C=45°, incorrectly getting 6/sin(A) = 8/sin(B) instead of using c² = 36 + 64 - 96cos(45°) = 32.2
- When using law of cosines to find angles, students forget to use the rearranged formula cos(C) = (a²+b²-c²)/(2ab), instead trying c² = a² + b² - 2ab·cos(C) and getting stuck with the unknown angle
- Students mix up opposite sides and angles in sine rule, writing a/sin(B) = b/sin(A) instead of a/sin(A) = b/sin(B), leading to incorrect ratios like 10/sin(60°) = 8/sin(30°) giving 11.5 instead of 4.6