Answer to Question #151565 in Trigonometry for Cavajos Montenegro

Question

In solving triangles, when do you use Law of Sines? How about the Law of Cosines? Give at least 2 examples for each.

Answer 1

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)

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

2)

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

Solving with the Law of Cosines:

1)

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

2)

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