Answer to Question #151565 in Trigonometry for Cavajos Montenegro

Question #151565
In solving triangles, when do you use Law of Sines? How about the Law of Cosines? Give at least 2 examples for each.
1
Expert's answer
2020-12-20T17:59:27-0500

Use the Law of Sines if:

  • You know two sides and an angle that is not the angle between the sides.
  • You know two angles and one side.

Use the Law of Cosines if:

  • You know three sides.
  • You know two sides and an angle between them.


Examples:

Solving with the Law of Sines:

1)



  • "\\frac{AB}{sin\\angle{C}}=\\frac{AC}{sin\\angle{B}}"
  • "sin\\angle{B}=\\frac{sin\\angle{C}*AC}{AB} = \\frac{\\frac{1}{2}*3\\sqrt{3}}{3} =\\frac{\\sqrt{3}}{2}"
  • "\\angle{B} = arcsin(sin\\angle{B}) = arcsin(\\frac{\\sqrt{3}}{2}) = 60\u00b0"
  • "\\angle{A} + \\angle{B} + \\angle{C} = 180\u00b0"
  • "\\angle{A} = 180\u00b0-\\angle{B} - \\angle{C} = 180\u00b0-60\u00b0-30\u00b0 = 90\u00b0"
  • "\\frac{AB}{sin\\angle{C}}=\\frac{BC}{sin\\angle{A}}"
  • "BC=\\frac{AB*sin\\angle{A}}{sin\\angle{C}}=\\frac{3*1}{\\frac{1}{2}} = 6"
  • Triangle solved!

2)


  • "\\angle{A} + \\angle{B} + \\angle{C} = 180\u00b0"
  • "\\angle{B} = 180\u00b0-\\angle{A} - \\angle{C} = 180\u00b0-90\u00b0-30\u00b0 = 60\u00b0"
  • "\\frac{AB}{sin\\angle{C}}=\\frac{BC}{sin\\angle{A}}"
  • "AB=\\frac{BC*sin\\angle{C}}{sin\\angle{A}}=\\frac{6*\\frac{1}{2}}{1} = 3"
  • "\\frac{AC}{sin\\angle{B}}=\\frac{BC}{sin\\angle{A}}"
  • "AC=\\frac{BC*sin\\angle{B}}{sin\\angle{A}}=\\frac{6*\\frac{\\sqrt{3}}{2}}{1} = 3\\sqrt{3}"
  • Triangle solved!


Solving with the Law of Cosines:

1)


  • "BC^2=AB^2+AC^2-2*AB*AC*cos\\angle{A}"
  • "cos\\angle{A} = \\frac{BC^2-AB^2-AC^2}{2*AB*AC} = \\frac{6^2-3^2-(3\\sqrt{3})^2}{2*3\\sqrt{3}} = 0"
  • "\\angle{A} = arccos(cos\\angle{A}) = arccos(0) = 90\u00b0"
  • "AB^2=BC^2+AC^2-2*BC*AC*cos\\angle{C}"
  • "cos\\angle{C} = \\frac{AB^2-BC^2-AC^2}{2*BC*AC} = \\frac{3^2-6^2-(3\\sqrt{3})^2}{2*6*3\\sqrt{3}} = \\frac{\\sqrt{3}}{2}"
  • "\\angle{C} = arccos(cos\\angle{C}) = arccos(\\frac{\\sqrt{3}}{2}) = 30\u00b0"
  • "\\angle{A}+\\angle{B}+\\angle{C} = 180\u00b0"
  • "\\angle{B}= 180\u00b0-\\angle{A}-\\angle{C} = 180\u00b0-90\u00b0-30\u00b0 = 60\u00b0"
  • Triangle solved!

2)

  • "BC^2=AB^2+AC^2-2*AB*AC*cos\\angle{A}"
  • "BC = \\sqrt{AB^2+AC^2-2*AB*AC*cos\\angle{A}} = \\sqrt{3^2+(3\\sqrt{3})^2-2*3*3\\sqrt{3}*0}=\\sqrt{36}=6"
  • "AB^2=BC^2+AC^2-2*BC*AC*cos\\angle{C}"
  • "cos\\angle{C} = \\frac{AB^2-BC^2-AC^2}{2*BC*AC} = \\frac{3^2-6^2-(3\\sqrt{3})^2}{2*6*3\\sqrt{3}} = \\frac{\\sqrt{3}}{2}"
  • "\\angle{C} = arccos(cos\\angle{C}) = arccos(\\frac{\\sqrt{3}}{2}) = 30\u00b0"
  • "\\angle{A}+\\angle{B}+\\angle{C} = 180\u00b0"
  • "\\angle{B}= 180\u00b0-\\angle{A}-\\angle{C} = 180\u00b0-90\u00b0-30\u00b0 = 60\u00b0"
  • Triangle solved!

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