Study Notes on the Law of Cosines and Application to Triangles

Law of Cosines

Introduction to the Law of Cosines

  • The Law of Cosines is a mathematical formula used to relate the lengths of the sides of a triangle with the cosine of one of its angles. It is particularly useful in solving triangles when specific combinations of sides and angles are known.

Cases for Using Law of Cosines

  • Case 3: SAS (Side-Angle-Side)

    • When two sides and the included angle are known:

    • The formula is as follows:

      • c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab \cos(C)

  • Case 4: SSS (Side-Side-Side)

    • When all three sides are known:

    • The formulas are:

      • b2=a2+c2−2accos⁡(B)b^2 = a^2 + c^2 - 2ac \cos(B)

      • a2=b2+c2−2bccos⁡(A)a^2 = b^2 + c^2 - 2bc \cos(A)

Proof of the Law of Cosines

  • To prove the Law of Cosines, consider a triangle where sides are labeled as $a$, $b$, and $c$ opposite to angles $A$, $B$, and $C$, respectively.

    • Start with the right triangle formed by dropping a perpendicular from point A to side BC, creating two segments on BC.

    • The Law of Cosines can be expressed in terms of the coordinates and distances within the triangle.

    • The formula can be derived using the Pythagorean theorem:

    • c2=(b−acos⁡(C))2+(asin⁡(C))2c^2 = (b - a \cos(C))^2 + (a \sin(C))^2

    • Simplifying this using the identity cos⁡2(C)+sin⁡2(C)=1\cos^2(C) + \sin^2(C) = 1 leads to:

    • c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab \cos(C)

Example Problems

Example 1: Solve the Triangle with SAS (a = 2, b = 3, C = 60°)

  • Given:

    • Sides a and b, and angle C.

  • Use the SAS formula:

    • c2=a2+b2−2abcos⁡(C)c^2 = a^2 + b^2 - 2ab \cos(C)

    • Substitute known values:

    • c2=22+32−2×2×3×cos⁡(60°)c^2 = 2^2 + 3^2 - 2 \times 2 \times 3 \times \cos(60°)

    • Simplify:

      • c2=4+9−2×2×3×12c^2 = 4 + 9 - 2 \times 2 \times 3 \times \frac{1}{2}

      • c2=13−6=7c^2 = 13 - 6 = 7

    • Therefore, c=7c = \sqrt{7}.

Example 2: Solve the Triangle with SSS (a = 4, b = 3, C = unknown)

  • Given:

    • Sides a, b, and calculated angle C.

  • Step 1 - Find angle A:

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

    • Substitute values:

    • cos⁡(A)=32+62−422×3×6\cos(A) = \frac{3^2 + 6^2 - 4^2}{2 \times 3 \times 6}

    • cos⁡(A)=9+36−1636\cos(A) = \frac{9 + 36 - 16}{36}

    • cos⁡(A)=2936\cos(A) = \frac{29}{36}

    • Compute angle A.

Example 3: Application Problem

  • Scenario: A motorized sailboat leaves Naples, Florida, bound for Key West, 150 miles away, traveling at a speed of 15 mph.

  • Given:

    • After 4 hours, due to crosswinds, the map course is off by 20%.

    • Calculate the following:

    • a) The distance from Key West after the 4 hours:

      • Adjusted distance:

      • d2=1502+602−2(150)(60)cos⁡(20°)d^2 = 150^2 + 60^2 - 2(150)(60) \cos(20°)

      • This gives approx. 295.8 miles.

    • b) What angle should the sailboat turn to correct its course?

      • Using:

      • 1502=962+602−2(96)(60)cos⁡(A)150^2 = 96^2 + 60^2 - 2(96)(60) \cos(A)

      • Solving for angle A gives A=147.2°A = 147.2° leading to an adjustment angle of 180°−A180° - A for correction.

    • c) Time added to the trip due to delay:

      • Total travel adjusted is 96+60=15696 + 60 = 156 miles.

      • Total time to travel this distance at 15 mph is 10.4 hours, thus an extra 24 minutes added to the original time.