Trigonometry: Sine Rule, Cosine Rule and Triangle Area

Statement

• a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius of the triangle.
• Use it when you know a matching pair: a side and the angle opposite to it.

Worked examples

1. Given a = 7, A = 40°, B = 60°: b = a·sin B / sin A = 7·sin 60° / sin 40° = 7(0.866)/0.643 ≈ 9.43.
2. Given b = 10, B = 50°, C = 70°: c = 10·sin 70° / sin 50° = 10(0.940)/0.766 ≈ 12.27.
3. Finding a side with the circumradius: a = 2R sin A. If R = 6 and A = 30°, then a = 12 × 0.5 = 6.

The ambiguous case

• When you are given two sides and a non-included angle (SSA), two different triangles may satisfy the data.
• Because sin(180° − A) = sin A, both A and 180° − A are possible; check whether both angles leave a valid third angle.
• Example: a = 6, b = 8, A = 35° → sin B = 8 sin 35°/6 ≈ 0.765 → B ≈ 49.9° or B ≈ 130.1°. Both are valid because 49.9° + 35° < 180° and 130.1° + 35° < 180°.

Statement

• a² = b² + c² − 2bc cos A.
• For an angle: cos A = (b² + c² − a²) / (2bc).
• Use it when you know two sides and the included angle (SAS) or all three sides (SSS).

Worked examples

1. Sides b = 5, c = 7, included angle A = 60°: a² = 25 + 49 − 2(5)(7)(0.5) = 74 − 35 = 39 → a = √39 ≈ 6.24.
2. Sides a = 8, b = 6, c = 5; find angle A: cos A = (36 + 25 − 64)/(2·6·5) = −3/60 = −0.05 → A ≈ 92.9°.
3. A right angle check: if a² = b² + c², then cos A = 0 and A = 90° — Pythagoras is the cosine rule with cos 90° = 0.

Choosing between the rules

The formula

• Area = ½ ab sin C, where a and b are two sides and C is the angle between them.
• Example: a = 5, b = 7, C = 60° → Area = ½(5)(7)(sin 60°) = 17.5 × 0.866 ≈ 15.16.
• Example: b = 10, c = 12, A = 30° → Area = ½(10)(12)(0.5) = 30.
• Heron's formula for three sides: Area = √(s(s − a)(s − b)(s − c)), where s = (a + b + c)/2.
• Example: sides 5, 6, 7 → s = 9 → Area = √(9·4·3·2) = √216 ≈ 14.70.

Choosing the right height

• The classic ½ × base × height is still valid; use the trigonometric form when the height is not given.
• In an obtuse triangle, sin C stays positive, so the formula still gives a positive area.

1. a = 9, A = 45°, B = 30°; find b → b = 9 sin30/sin45 = 6.36
2. b = 12, c = 15, A = 120°; find a → a² = 144 + 225 − 360(−0.5) = 549 → a ≈ 23.43
3. Sides 7, 8, 10; find the largest angle → cos C = (49 + 64 − 100)/(2·7·8) = 13/112 → C ≈ 83.3°
4. Two sides 6 and 8 with included angle 45°; find the area → ½·6·8·sin45 = 16.97
5. In a triangle, a = 4, b = 6, A = 30°; how many triangles are possible? → sin B = 6 sin30/4 = 0.75 → B ≈ 48.6° or 131.4°; only the first leaves C > 0 with the second? 30 + 131.4 < 180 as well — check both and state both; the safe answer is two possibilities
6. A triangle has area 30 and sides 10 and 12; find the included angle → 30 = ½·10·12·sin C → sin C = 0.5 → C = 30° or 150°