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Law of Cosines

Definition and Fundamental Concepts

  • The Law of Cosines is a mathematical formula used to find unknown sides or angles in triangles.

  • The formulas are defined as follows:

    • a2=b2+c2−2bc extcos Aa^2 = b^2 + c^2 - 2bc \, ext{cos} \ A

    • b2=a2+c2−2ac extcos Bb^2 = a^2 + c^2 - 2ac \, ext{cos} \ B

    • c2=a2+b2−2ab extcos Cc^2 = a^2 + b^2 - 2ab \, ext{cos} \ C

Scenarios for Law of Sines vs. Law of Cosines

  • The Law of Sines is applicable in specific cases, including:

    • ASA (Angle-Side-Angle)

    • SAA (Side-Angle-Angle)

    • Two angles and the included side

    • Two angles and any side

    • SSA (Side-Side-Angle, noting possible ambiguity)

    • Two sides and an opposite angle (this case has an awareness of possible ambiguity)

Why Law of Sines Cannot Work for SAS Triangle
  • In an SAS triangle setup, you are given two sides and the included angle. The reason the Law of Sines is not applicable:

    • There is no side opposite the angle of interest, preventing the establishment of a ratio necessary for using the Law of Sines.

Utilization of Law of Cosines

Right Triangles
  • When angle CC is exactly 90 degrees, (i.e., ZC=90°ZC = 90°), we can simplify the Law of Cosines to:

    • c2=a2+b2c^2 = a^2 + b^2

    • This is the Pythagorean theorem.

Acute Triangles
  • When angle CC is less than 90 degrees (i.e., ZC<90°ZC < 90°), the cosine function is positive:

    • c2<a2+b2c^2 < a^2 + b^2 by an amount of 2ab extcos C2ab \, ext{cos} \ C.

Obtuse Triangles
  • When angle CC is greater than 90 degrees (i.e., ZC>90°ZC > 90°), the cosine function is negative:

    • c2>a2+b2c^2 > a^2 + b^2 by an amount of −2ab extcos C-2ab \, ext{cos} \ C.

Finding Missing Sides and Angles

  • Rearranging the Law of Cosines to find angles:

    • cos A=b2+c2−a22bc\text{cos} \ A = \frac{b^2 + c^2 - a^2}{2bc}

    • cos B=a2+c2−b22ac\text{cos} \ B = \frac{a^2 + c^2 - b^2}{2ac}

    • cos C=a2+b2−c22ab\text{cos} \ C = \frac{a^2 + b^2 - c^2}{2ab}

  • This rearranged form allows for angle calculation if the lengths of all three sides are known.

  • The general structure for this form is:

    • cos (angle)=(adjacent)2+(adjacent)2−(opposite)22(adjacent)(adjacent)\text{cos} \ (\text{angle}) = \frac{(\text{adjacent})^2 + (\text{adjacent})^2 - (\text{opposite})^2}{2(\text{adjacent})(\text{adjacent})}

Choosing the Method to Solve Oblique Triangles

  • Different methods are suitable based on the information given:

    • SAA or ASA: Use Law of Sines (opposite the given side).

    • SSA (Ambiguous Case): Use Law of Sines.

    • SAS: Use Law of Cosines.

    • SSS: Use Law of Cosines.

Example 1: Solving Triangle with Known Side Lengths and Angles

Given:
  • Angle A=30°A = 30°

  • Side b=4b = 4

  • Side c=5c = 5

Finding Side aa Using Law of Cosines:
  • Formulation:

    • a2=b2+c2−2bc extcos Aa^2 = b^2 + c^2 - 2bc \, ext{cos} \ A

    • Substituting values:

    • a2=42+52−2(4)(5) extcos 30°a^2 = 4^2 + 5^2 - 2(4)(5) \, ext{cos} \ 30°

    • Calculate:

      • a2=16+25−2(4)(5)(0.866)a^2 = 16 + 25 - 2(4)(5)(0.866) (approx 6.36)

    • Result: a≈2.52a \approx 2.52

Finding Angle BB Using Law of Cosines:
  • Formulation:

    • b2=c2+a2−2ca extcos Bb^2 = c^2 + a^2 - 2ca \, ext{cos} \ B

    • Substituting results:

    • b2=52+(2.52)2−2(5)(2.52) extcos Bb^2 = 5^2 + (2.52)^2 - 2(5)(2.52) \, ext{cos} \ B

  • Calculating yields: 0.60910.6091, i.e.,

    • cos B=52+(2.52)2−422(5)(2.52)  ⟹  B=52.5°\text{cos} \ B = \frac{5^2 + (2.52)^2 - 4^2}{2(5)(2.52)} \implies B = 52.5°

Finding Angle CC:
  • Result: C=180°−A−B=180°−30°−52.5°=97.5°C = 180° - A - B = 180° - 30° - 52.5° = 97.5°

Example 2: Solving with Sides A = 2, B = 3, C = 4

Given:
  • Side lengths a=2a = 2, b=3b = 3, and c=4c = 4

Finding Angle BB Using Law of Cosines:
  • Calculation:

    • b2=c2+a2−2ca cos Bb^2 = c^2 + a^2 - 2ca \, \text{cos} \ B

    • Substituting:

    • 32=42+22−2(4)(2) cos B3^2 = 4^2 + 2^2 - 2(4)(2) \, \text{cos} \ B

    • Result yields: B=46.6°B = 46.6°

Finding Angle CC Using Law of Cosines:
  • Formulation:

    • c2=a2+b2−2ab cos Cc^2 = a^2 + b^2 - 2ab \, \text{cos} \ C

  • Substituting:

    • Find CC yields: 104.5°104.5°

Finding Angle AA:
  • Calculate: A=180°−B−C=28.9°A = 180° - B - C = 28.9°

Example 3: No Valid Triangle with Sides A = 2, B = 3, C = 6

Given:
  • Side lengths a=2a = 2, b=3b = 3, and c=6c = 6

  • Observing the triangle inequality:

    • Recognize immediately:

    • The length of side cc is greater than the sum of side lengths 2+32 + 3, hence,

    • This scenario is impossible for a triangle.

    • Calculation attempts will yield non-real results for angles.

Conclusion

  • The Law of Cosines is integral in finding unknowns in various triangle scenarios, especially for SAS and SSS cases, reinforcing the need for understanding its application in both acute and obtuse triangle properties. The understanding of triangle inequality is crucial for establishing possible configurations of side lengths.