8.2Solving Oblique Triangles with the Law of Cosines
Apply c² = a² + b² − 2ab cosC to solve SAS and SSS triangles. Use Heron's formula A = √(s(s−a)(s−b)(s−c)) to find area when all three sides are known.
The Law of Cosines generalizes the Pythagorean theorem to any triangle. It is used in navigation, architecture, and physics whenever three sides or two sides and the included angle are known.
Essential Question
When you know two sides and the included angle (SAS) or all three sides (SSS) of a triangle, how do you find the remaining measurements?
Lesson Overview
Law of Cosines — Three Equivalent Forms
a² = b² + c² − 2bc cosA
b² = a² + c² − 2ac cosB
c² = a² + b² − 2ab cosC
Solving for an Angle
cosC = (a² + b² − c²) / (2ab)
Similarly for cosA and cosB by cycling the letters.
When to Use
• SAS — two sides and the included angle are known
• SSS — all three sides are known
Area — Heron's Formula
s = (a + b + c) / 2
Area = √(s(s−a)(s−b)(s−c))
Decision Tree
• AAS / ASA → Law of Sines
• SAS / SSS → Law of Cosines
• SSA → Law of Sines (check for ambiguous case)
Worked Examples
SAS: a = 10, b = 14, C = 52°. Find side c, then angles A and B.
Apply c² = a² + b² − 2ab cosC
c² = 10² + 14² − 2(10)(14)cos52°
c² = 100 + 196 − 280(0.6157)
c² = 296 − 172.4 = 123.6
c ≈ 11.12
Now use Law of Sines: sinA/a = sinC/c
sinA = 10 · sin52° / 11.12 = 10(0.7880)/11.12 ≈ 0.7086
A ≈ 45.0°
B = 180° − 52° − 45.0° = 83.0°
SSS: a = 7, b = 9, c = 12. Find all three angles.
Find the largest angle first (opposite longest side c = 12):
cosC = (a² + b² − c²) / (2ab) = (49 + 81 − 144) / (2·7·9)
cosC = −14 / 126 ≈ −0.1111 → C ≈ 96.4°
Now use Law of Sines for A:
sinA = a · sinC / c = 7 · sin96.4° / 12 = 7(0.9938)/12 ≈ 0.5797
A ≈ 35.4°
B = 180° − 96.4° − 35.4° = 48.2°
Heron's formula: sides 8, 11, 15. Find the area.
s = (8 + 11 + 15) / 2 = 34 / 2 = 17
Area = √(s(s−a)(s−b)(s−c))
Area = √(17 · 9 · 6 · 2)
Area = √(1836)
Area ≈ 42.8 square units
Navigation: Two ships leave port. One travels 30 km at bearing N 40° E, the other 45 km at bearing N 70° W. Find the distance between them.
The angle between the two bearings: 40° (east of north) + 70° (west of north) = 110°
This is the included angle C between sides a = 45 km and b = 30 km.
c² = a² + b² − 2ab cosC
c² = 45² + 30² − 2(45)(30)cos110°
c² = 2025 + 900 − 2700(−0.3420)
c² = 2925 + 923.4 = 3848.4
c ≈ 62.0 km
SAS: b = 20, c = 25, A = 110°. Find side a.
Apply a² = b² + c² − 2bc cosA
a² = 20² + 25² − 2(20)(25)cos110°
a² = 400 + 625 − 1000(−0.3420)
a² = 1025 + 342.0 = 1367.0
a ≈ 36.97
Guided Practice
SAS: a = 6, c = 9, B = 70°. Find side b.
Hint: Use b² = a² + c² − 2ac cosB, then take the square root.
SSS: a = 5, b = 7, c = 8. Find angle C.
Hint: Use cosC = (a² + b² − c²) / (2ab). Substitute and evaluate with a calculator.
Heron's formula: sides 13, 14, 15. Find the area.
Hint: Compute s = (13+14+15)/2 = 21, then Area = √(21·8·7·6).
SAS: two sides 18 and 24, included angle 35°. Find the third side and the area.
Hint: Use the Law of Cosines for the third side, then Area = (1/2)ab sinC for the area.
A triangular plot has sides 120 m, 150 m, 200 m. Find the largest angle.
Hint: The largest angle is opposite the longest side (200 m). Use cosC = (a²+b²−c²)/(2ab).
Key Vocabulary
Law of Cosines
A formula relating the three sides and one angle of any triangle: c² = a² + b² − 2ab cosC.
Example: For a = 5, b = 7, C = 60°: c² = 25 + 49 − 70(0.5) = 39, so c ≈ 6.24
SAS (Side-Angle-Side)
A triangle configuration where two sides and the included angle between them are known.
Example: Sides 8 and 11 with the angle between them equal to 50°.
SSS (Side-Side-Side)
A triangle configuration where all three side lengths are known.
Example: Sides 6, 8, and 10 — a right triangle.
Heron's Formula
A formula for the area of a triangle given all three sides: Area = √(s(s−a)(s−b)(s−c)).
Example: Sides 3, 4, 5: s = 6, Area = √(6·3·2·1) = √36 = 6
Semi-perimeter
Half the perimeter of a triangle: s = (a + b + c) / 2. Used in Heron's formula.
Example: For sides 5, 7, 8: s = 20/2 = 10
Included angle
The angle formed between two known sides of a triangle.
Example: In SAS with sides b and c, angle A is the included angle.
Oblique triangle
Any triangle that is not a right triangle (no 90° angle).
Example: A triangle with angles 52°, 83°, and 45° is oblique.
Check Your Understanding
Interactive Practice — 5 Questions
The Law of Cosines is used for which triangle cases?
In the formula cosC = (a²+b²−c²)/(2ab), if the result is negative, angle C is:
Heron's formula requires:
For a = 5, b = 7, C = 60°, c² = ?
The semi-perimeter of a triangle with sides 6, 8, 10 is:
Common Mistakes
Using the Law of Cosines for AAS or ASA triangles.
AAS and ASA give you two angles and a side — use the Law of Sines, which is simpler and correct for those cases.
Stopping after computing c² = … and reporting c² as the answer.
Always take the square root: c = √(c²). The Law of Cosines gives you c², not c.
Writing cosC = (a²+b²+c²)/(2ab) — adding c² instead of subtracting.
The correct form is cosC = (a²+b²−c²)/(2ab). The c² term is subtracted.
In Heron's formula, computing s = a+b+c instead of (a+b+c)/2.
s is the semi-perimeter: s = (a+b+c)/2. Forgetting to divide by 2 gives a wildly incorrect area.
Math Tips
The Law of Cosines reduces to the Pythagorean theorem when C = 90°: c² = a² + b² − 2ab·0 = a² + b².
After finding one angle with the Law of Cosines, use the Law of Sines for the remaining angles — it's simpler.
Always find the largest angle first in SSS — it's opposite the longest side and tells you if the triangle is acute or obtuse.
Heron's formula is exact and elegant but sensitive to rounding. Keep full precision in s until the final step.
Decision rule: count what you know. Two sides + included angle = SAS → Law of Cosines. Three sides = SSS → Law of Cosines.