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)
A B s i n ∠ C = A C s i n ∠ B \frac{AB}{sin\angle{C}}=\frac{AC}{sin\angle{B}} s in ∠ C A B = s in ∠ B A C s i n ∠ B = s i n ∠ C ∗ A C A B = 1 2 ∗ 3 3 3 = 3 2 sin\angle{B}=\frac{sin\angle{C}*AC}{AB} = \frac{\frac{1}{2}*3\sqrt{3}}{3} =\frac{\sqrt{3}}{2} s in ∠ B = A B s in ∠ C ∗ A C = 3 2 1 ∗ 3 3 = 2 3 ∠ B = a r c s i n ( s i n ∠ B ) = a r c s i n ( 3 2 ) = 60 ° \angle{B} = arcsin(sin\angle{B}) = arcsin(\frac{\sqrt{3}}{2}) = 60° ∠ B = a rcs in ( s in ∠ B ) = a rcs in ( 2 3 ) = 60° ∠ A + ∠ B + ∠ C = 180 ° \angle{A} + \angle{B} + \angle{C} = 180° ∠ A + ∠ B + ∠ C = 180° ∠ A = 180 ° − ∠ B − ∠ C = 180 ° − 60 ° − 30 ° = 90 ° \angle{A} = 180°-\angle{B} - \angle{C} = 180°-60°-30° = 90° ∠ A = 180° − ∠ B − ∠ C = 180° − 60° − 30° = 90° A B s i n ∠ C = B C s i n ∠ A \frac{AB}{sin\angle{C}}=\frac{BC}{sin\angle{A}} s in ∠ C A B = s in ∠ A BC B C = A B ∗ s i n ∠ A s i n ∠ C = 3 ∗ 1 1 2 = 6 BC=\frac{AB*sin\angle{A}}{sin\angle{C}}=\frac{3*1}{\frac{1}{2}} = 6 BC = s in ∠ C A B ∗ s in ∠ A = 2 1 3 ∗ 1 = 6 Triangle solved! 2)
∠ A + ∠ B + ∠ C = 180 ° \angle{A} + \angle{B} + \angle{C} = 180° ∠ A + ∠ B + ∠ C = 180° ∠ B = 180 ° − ∠ A − ∠ C = 180 ° − 90 ° − 30 ° = 60 ° \angle{B} = 180°-\angle{A} - \angle{C} = 180°-90°-30° = 60° ∠ B = 180° − ∠ A − ∠ C = 180° − 90° − 30° = 60° A B s i n ∠ C = B C s i n ∠ A \frac{AB}{sin\angle{C}}=\frac{BC}{sin\angle{A}} s in ∠ C A B = s in ∠ A BC A B = B C ∗ s i n ∠ C s i n ∠ A = 6 ∗ 1 2 1 = 3 AB=\frac{BC*sin\angle{C}}{sin\angle{A}}=\frac{6*\frac{1}{2}}{1} = 3 A B = s in ∠ A BC ∗ s in ∠ C = 1 6 ∗ 2 1 = 3 A C s i n ∠ B = B C s i n ∠ A \frac{AC}{sin\angle{B}}=\frac{BC}{sin\angle{A}} s in ∠ B A C = s in ∠ A BC A C = B C ∗ s i n ∠ B s i n ∠ A = 6 ∗ 3 2 1 = 3 3 AC=\frac{BC*sin\angle{B}}{sin\angle{A}}=\frac{6*\frac{\sqrt{3}}{2}}{1} = 3\sqrt{3} A C = s in ∠ A BC ∗ s in ∠ B = 1 6 ∗ 2 3 = 3 3 Triangle solved!
Solving with the Law of Cosines:
1)
B C 2 = A B 2 + A C 2 − 2 ∗ A B ∗ A C ∗ c o s ∠ A BC^2=AB^2+AC^2-2*AB*AC*cos\angle{A} B C 2 = A B 2 + A C 2 − 2 ∗ A B ∗ A C ∗ cos ∠ A c o s ∠ A = B C 2 − A B 2 − A C 2 2 ∗ A B ∗ A C = 6 2 − 3 2 − ( 3 3 ) 2 2 ∗ 3 3 = 0 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 cos ∠ A = 2 ∗ A B ∗ A C B C 2 − A B 2 − A C 2 = 2 ∗ 3 3 6 2 − 3 2 − ( 3 3 ) 2 = 0 ∠ A = a r c c o s ( c o s ∠ A ) = a r c c o s ( 0 ) = 90 ° \angle{A} = arccos(cos\angle{A}) = arccos(0) = 90° ∠ A = a rccos ( cos ∠ A ) = a rccos ( 0 ) = 90° A B 2 = B C 2 + A C 2 − 2 ∗ B C ∗ A C ∗ c o s ∠ C AB^2=BC^2+AC^2-2*BC*AC*cos\angle{C} A B 2 = B C 2 + A C 2 − 2 ∗ BC ∗ A C ∗ cos ∠ C c o s ∠ C = A B 2 − B C 2 − A C 2 2 ∗ B C ∗ A C = 3 2 − 6 2 − ( 3 3 ) 2 2 ∗ 6 ∗ 3 3 = 3 2 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} cos ∠ C = 2 ∗ BC ∗ A C A B 2 − B C 2 − A C 2 = 2 ∗ 6 ∗ 3 3 3 2 − 6 2 − ( 3 3 ) 2 = 2 3 ∠ C = a r c c o s ( c o s ∠ C ) = a r c c o s ( 3 2 ) = 30 ° \angle{C} = arccos(cos\angle{C}) = arccos(\frac{\sqrt{3}}{2}) = 30° ∠ C = a rccos ( cos ∠ C ) = a rccos ( 2 3 ) = 30° ∠ A + ∠ B + ∠ C = 180 ° \angle{A}+\angle{B}+\angle{C} = 180° ∠ A + ∠ B + ∠ C = 180° ∠ B = 180 ° − ∠ A − ∠ C = 180 ° − 90 ° − 30 ° = 60 ° \angle{B}= 180°-\angle{A}-\angle{C} = 180°-90°-30° = 60° ∠ B = 180° − ∠ A − ∠ C = 180° − 90° − 30° = 60° Triangle solved! 2)
B C 2 = A B 2 + A C 2 − 2 ∗ A B ∗ A C ∗ c o s ∠ A BC^2=AB^2+AC^2-2*AB*AC*cos\angle{A} B C 2 = A B 2 + A C 2 − 2 ∗ A B ∗ A C ∗ cos ∠ A B C = A B 2 + A C 2 − 2 ∗ A B ∗ A C ∗ c o s ∠ A = 3 2 + ( 3 3 ) 2 − 2 ∗ 3 ∗ 3 3 ∗ 0 = 36 = 6 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 BC = A B 2 + A C 2 − 2 ∗ A B ∗ A C ∗ cos ∠ A = 3 2 + ( 3 3 ) 2 − 2 ∗ 3 ∗ 3 3 ∗ 0 = 36 = 6 A B 2 = B C 2 + A C 2 − 2 ∗ B C ∗ A C ∗ c o s ∠ C AB^2=BC^2+AC^2-2*BC*AC*cos\angle{C} A B 2 = B C 2 + A C 2 − 2 ∗ BC ∗ A C ∗ cos ∠ C c o s ∠ C = A B 2 − B C 2 − A C 2 2 ∗ B C ∗ A C = 3 2 − 6 2 − ( 3 3 ) 2 2 ∗ 6 ∗ 3 3 = 3 2 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} cos ∠ C = 2 ∗ BC ∗ A C A B 2 − B C 2 − A C 2 = 2 ∗ 6 ∗ 3 3 3 2 − 6 2 − ( 3 3 ) 2 = 2 3 ∠ C = a r c c o s ( c o s ∠ C ) = a r c c o s ( 3 2 ) = 30 ° \angle{C} = arccos(cos\angle{C}) = arccos(\frac{\sqrt{3}}{2}) = 30° ∠ C = a rccos ( cos ∠ C ) = a rccos ( 2 3 ) = 30° ∠ A + ∠ B + ∠ C = 180 ° \angle{A}+\angle{B}+\angle{C} = 180° ∠ A + ∠ B + ∠ C = 180° ∠ B = 180 ° − ∠ A − ∠ C = 180 ° − 90 ° − 30 ° = 60 ° \angle{B}= 180°-\angle{A}-\angle{C} = 180°-90°-30° = 60° ∠ B = 180° − ∠ A − ∠ C = 180° − 90° − 30° = 60° Triangle solved! 
#339914 on Dec 2023
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