Video summary

Trigonometri • Part 26: Contoh Soal Aturan Sinus, Aturan Cosinus & Luas Segitiga (1)

Main summary

Key takeaways

Educational

Main Ideas / Concepts

  • The video teaches trigonometry example problems involving:
    • Aturan Sinus (Law of Sines)
    • Aturan Cosinus (Law of Cosines) (mentioned as a topic, though the worked examples shown focus on the sine rule and area)
    • Luas segitiga (Area of a triangle), including Heron’s formula
  • It also demonstrates how to find the area of a regular hexagon by decomposing it into congruent triangles.

Methodology / Instructions (Step-by-Step)

1) Example: Find side (AB) using the Law of Sines

Given:

  • Triangle (ABC)
  • (BC = 6\sqrt{2}) cm
  • (\angle BAC = 60^\circ)
  • (\angle ACB = 45^\circ)
  • Find (AB)

Steps:

  1. Identify what’s known: two angles and one side. The Law of Sines is appropriate because it connects each side with the sine of its opposite angle.

  2. Use the Law of Sines: [ \frac{\text{side}}{\sin(\text{opposite angle})} = \frac{BC}{\sin(\angle B)} = \frac{AB}{\sin(\angle C)} ]

  3. Match known side to its opposite angle:

    • (BC) is opposite (\angle A = 60^\circ)
  4. Match (AB) to its opposite angle:
    • (AB) is opposite (\angle C = 45^\circ)

  5. Set up: [ \frac{BC}{\sin 60^\circ}=\frac{AB}{\sin 45^\circ} ]

  6. Substitute values: [ \frac{6\sqrt{2}}{\sin 60^\circ}=\frac{AB}{\sin 45^\circ} ]

  7. Use trig values:

    • (\sin 60^\circ = \frac{\sqrt{3}}{2})
    • (\sin 45^\circ = \frac{\sqrt{2}}{2})

  8. Solve: [ AB = 6\sqrt{2}\cdot \frac{\sin 45^\circ}{\sin 60^\circ} ]

  9. Simplify to: [ AB = 4\sqrt{3}\text{ cm} ]

Result: [ AB = 4\sqrt{3}\text{ cm} ]

2) Example: Find the area of a triangle with sides 3 cm, 6 cm, 7 cm (Heron’s Formula)

Given:

  • Triangle sides: (a=3), (b=6), (c=7) (cm)
  • Find area

Steps (Heron’s formula):

  1. Compute the semiperimeter: [ s=\frac{a+b+c}{2} ]

  2. Substitute: [ s=\frac{3+6+7}{2}=\frac{16}{2}=8 ]

  3. Apply Heron’s formula: [ \text{Area}=\sqrt{s(s-a)(s-b)(s-c)} ]

  4. Substitute: [ \text{Area}=\sqrt{8(8-3)(8-6)(8-7)} ]

  5. Simplify inside the root: [ =\sqrt{8\cdot5\cdot2\cdot1} ]

  6. Continue simplifying: [ =\sqrt{80}=4\sqrt{5} ] Units are (\text{cm}^2).

Result: [ \text{Area}=4\sqrt{5}\text{ cm}^2 ]

3) Example: Find the area of a regular hexagon with side length 6 cm

Given:

  • Regular hexagon with side length (6) cm
  • Find area

Steps:

  1. Draw the regular hexagon.
  2. Decompose the hexagon into 6 congruent triangles by splitting from the center (or connecting opposite vertices).
  3. Each triangle is effectively an equilateral triangle with:
    • side length (6) cm
    • vertex angle (60^\circ)

  4. Compute the area of one equilateral triangle: [ \text{Area}=\frac{1}{2}\cdot \text{side}\cdot \text{side}\cdot \sin(60^\circ) ]

  5. Substitute side (=6): [ \text{Area}=\frac{1}{2}\cdot 6\cdot 6\cdot \sin 60^\circ ]

  6. Use (\sin 60^\circ=\frac{\sqrt{3}}{2}): [ \text{Area}=\frac{1}{2}\cdot 36\cdot \frac{\sqrt{3}}{2}=9\sqrt{3} ]

  7. Multiply by 6 congruent triangles: [ \text{Hexagon area}=6\cdot 9\sqrt{3}=54\sqrt{3} ]

Result: [ \text{Area of hexagon}=54\sqrt{3}\text{ cm}^2 ]

Speakers / Sources Featured

  • No specific individual speaker name is provided in the subtitles.
  • Source referenced: “Science Window” (channel name: Science Window channel)

Original video