Sections 4.3–4.4 Law of Cosines & Triangle Area (Comprehensive Study Notes)

Section 4.3 Law of Cosines

• Core formula (three equivalent versions – choose the one whose isolated letter is the unknown side):
• a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A
• b2=a2+c2−2accos⁡Bb^2=a^2+c^2-2ac\cos B
• c2=a2+b2−2abcos⁡Cc^2=a^2+b^2-2ab\cos C

• When do we use it?
• SAS (Side–Angle–Side): 2 sides and the included angle are known.
• SSS (Side–Side–Side): all 3 sides known – use the formula to find any angle.
• Avoid mixing with Law of Sines in the same problem unless directions explicitly say so – rounding in one law can corrupt answers in the other.

• Calculator reminders
• While solving for an angle, the numeric value of cos⁡(angle)\cos(\text{angle}) must lie between −1-1 and 11.
• Enter fractions carefully: putting an entire numerator in parentheses prevents the machine from treating     \sqrt{\;\;} as being in the denominator.
• Use four–decimal accuracy for cosine‐ratios before pressing cos⁡−1\cos^{-1}.

• Geometric fact-check
• Longest side is opposite the largest angle; shortest side opposite the smallest angle. A quick test for consistency when answers seem off.

Example 1 (SAS)

Triangle with sides b=2b=2, c=3c=3 and included angle C=60∘C=60^\circ.

  1. Find third side aa (called cc in the transcript but relabelled here for consistency)
    • a2=22+32−2(2)(3)cos⁡60∘=4+9−12⋅12=13−6=7a^2=2^2+3^2-2(2)(3)\cos60^\circ=4+9-12\cdot\tfrac12=13-6=7
    • a=7 (≈2.646)a=\sqrt7\,(\approx2.646)

  2. Find AA (opposite side aa)
    • Plug into cos⁡A=b2+c2−a22bc=4+9−72(2)(3)=612=0.7559\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{4+9-7}{2(2)(3)}=\dfrac{6}{12}=0.7559
    • A=cos⁡−1(0.7559)≈40.9∘A=\cos^{-1}(0.7559)\approx40.9^\circ

  3. Find BB by triangle-sum: B=180∘−60∘−40.9∘≈79.1∘B=180^\circ-60^\circ-40.9^\circ\approx79.1^\circ

Example 2 (SSS)

Triangle a=4a=4, b=3b=3, c=6c=6.

  1. Angle AA
    cos⁡A=42+62−322(4)(6)=16+36−948=0.8056→A≈36.3∘\cos A=\dfrac{4^2+6^2-3^2}{2(4)(6)}=\dfrac{16+36-9}{48}=0.8056\to A\approx36.3^\circ

  2. Angle BB
    cos⁡B=32+62−422(3)(6)=0.8958→B≈26.4∘\cos B=\dfrac{3^2+6^2-4^2}{2(3)(6)}=0.8958\to B\approx26.4^\circ

  3. C=180∘−36.3∘−26.4∘≈117.3∘C=180^\circ-36.3^\circ-26.4^\circ\approx117.3^\circ ✔ largest angle opposite longest side.

Application 1 (air-route)

Fort Myers → Sarasota = 150 mi150\text{ mi} (due north). Sarasota → Orlando = 100 mi100\text{ mi} at a 50∘50^\circ turn.

• Interior angle at Sarasota used in the triangle is 180∘−50∘=130∘180^\circ-50^\circ=130^\circ.

• Direct distance dd between Fort Myers and Orlando:
d2=1502+1002−2(150)(100)cos⁡130∘d^2=150^2+100^2-2(150)(100)\cos130^\circ
d≈227.6 mid\approx227.6\text{ mi}

Application 2 (Baseball diamond)

90 ft square; centre-back-stop fence 400 ft from home; find distance from fence to 3rd base.

• Interior angle at 2nd base is 45∘45^\circ.
• d2=4002+902−2(400)(90)cos⁡45∘→d≈342.3 ftd^2=400^2+90^2-2(400)(90)\cos45^\circ\to d\approx342.3\text{ ft}

Section 4.4 Area of a Triangle

  1. SAS area formula (derived from 12absin⁡C\tfrac12ab\sin C):
    Area=12 absin⁡C\boxed{\text{Area}=\tfrac12\,ab\sin C}

  2. SSS (Heron’s) formula
    • First compute s=a+b+c2s=\dfrac{a+b+c}{2}.
    • Then Area=s (s−a)(s−b)(s−c)\text{Area}=\sqrt{s\,(s-a)(s-b)(s-c)}.

Example 1 (SAS)

Sides 8, 6 with included angle 30∘30^\circ.
Area=12(8)(6)sin⁡30∘=24⋅12=12\text{Area}=\tfrac12(8)(6)\sin30^\circ=24\cdot\tfrac12=12 (square units).

Example 2 (Home-plate pentagon)

• Rectangle part: 17 in×8.5 in=144.5 in217\text{ in}\times8.5\text{ in}=144.5\text{ in}^2
• Triangular part with sides 12,12,17:
• s=12+12+172=20.5s=\tfrac{12+12+17}{2}=20.5
• Area =20.5 (8.5)(8.5)(3.5)≈71.99 in2=\sqrt{20.5\,(8.5)(8.5)(3.5)}\approx71.99\text{ in}^2
• Total ≈ 216.49 in2216.49\text{ in}^2.

Example 3 (Lake outline – composite)

• Two SAS triangles plus one interior SSS triangle.
• Upper triangle: 362.35 ft2362.35\text{ ft}^2
• Lower triangle: 443.16 ft2443.16\text{ ft}^2
• Central triangle: sides ~47.0747.07, 45.9645.96, 4040 → area 839.91 ft2839.91\text{ ft}^2
• Total lake area ≈ 1645.42 ft21645.42\text{ ft}^2.

Practical & Test Hints

• Always tag each side with its opposite angle before writing formulas; keeps the substitutions consistent.
• If your computed cos⁡(θ)∉[−1,1]\cos(\theta)\notin[-1,1], re-check algebra & parentheses.
• Heron’s formula is numerically stable; prefer it over ‘split-into-two-right-triangles’ when only sides are given.
• Big-picture roadmap for assessments (given by instructor):
• Quiz #4 covers Sections 4.1–4.4, 5.1, 5.3.
• Test #3 next week, same sections (possible 5.2 if reached).
• Final review distributed next Wednesday; homework accepted until the first week of May.

• Chapter 5 (Trigonometric Functions) begins Thursday; Section 5.3 is notably challenging—preview ahead if possible.