Lingkaran luar segitiga adalah lingkaran yang berada di luar segitiga dan ketiga sisi segitiga menyentuh (keliling/garis luar) lingkaran.
Lingkaran O merupakan lingkaran luar pada segitiga ABC karena titik A, B, dan C berada di lingkaran O.
Untuk mengetahui jari-jari lingkaran kita bisa menganalisa segitiga dan lingkaran di samping.
Sudut ∠ A C B \displaystyle{ \angle A C B } ∠ A CB ∠ A O B \displaystyle{ \angle A O B } ∠ A OB ∠ A C B \displaystyle{ \angle A C B } ∠ A CB ∠ A O B \displaystyle{ \angle A O B } ∠ A OB A B \displaystyle{ A B } A B
∠ A O B = 2 ( ∠ A C B ) \displaystyle{ \angle A O B = 2 \left( \angle A C B \right) } ∠ A OB = 2 ( ∠ A CB )
A O \displaystyle{ A O } A O B O \displaystyle{ B O } BO A O = B O \displaystyle{ A O = B O } A O = BO ∠ O A B = ∠ O B A \displaystyle{ \angle O A B = \angle O B A } ∠ O A B = ∠ OB A 180 ° \displaystyle{ 180 \degree } 180°
∠ O A B + ∠ A O B + ∠ O B A = 180 ° ∠ O A B + 2 ( ∠ A C B ) + ∠ O A B = 180 ° 2 ( ∠ O A B ) + 2 ( ∠ A C B ) = 180 ° 2 ( ∠ O A B + ∠ A C B ) = 180 ° ∠ O A B + ∠ A C B = 90 ° \displaystyle{ \begin{aligned}\angle O A B + \angle A O B + \angle O B A = 180 \degree \\ \angle O A B + 2 \left( \angle A C B \right) + \angle O A B = 180 \degree \\ 2 \left( \angle O A B \right) + 2 \left( \angle A C B \right) = 180 \degree \\ 2 \left( \angle O A B + \angle A C B \right) = 180 \degree \\ \angle O A B + \angle A C B = 90 \degree\end{aligned} } ∠ O A B + ∠ A OB + ∠ OB A = 180° ∠ O A B + 2 ( ∠ A CB ) + ∠ O A B = 180° 2 ( ∠ O A B ) + 2 ( ∠ A CB ) = 180° 2 ( ∠ O A B + ∠ A CB ) = 180° ∠ O A B + ∠ A CB = 90°
Berlaku hukum sinus pada segitiga AOB seperti berikut:
A O sin ( ∠ A B O ) = O B sin ( ∠ O A B ) = A B sin ( ∠ A O B ) r sin ( ∠ O A B ) = r sin ( ∠ O A B ) = A B sin ( ∠ A O B ) r sin ( ∠ O A B ) = A B sin ( ∠ A O B ) \displaystyle{ \begin{aligned}\frac{ A O }{ \sin \left( \angle A B O \right) } = & \frac{ O B }{ \sin \left( \angle O A B \right) } = \frac{ A B }{ \sin \left( \angle A O B \right) } \\ \frac{ r }{ \sin \left( \angle O A B \right) } = & \frac{ r }{ \sin \left( \angle O A B \right) } = \frac{ A B }{ \sin \left( \angle A O B \right) } \\ & \frac{ r }{ \sin \left( \angle O A B \right) } = \frac{ A B }{ \sin \left( \angle A O B \right) }\end{aligned} } sin ( ∠ A BO ) A O = sin ( ∠ O A B ) r = sin ( ∠ O A B ) OB = sin ( ∠ A OB ) A B sin ( ∠ O A B ) r = sin ( ∠ A OB ) A B sin ( ∠ O A B ) r = sin ( ∠ A OB ) A B
r = A B ⋅ sin ( ∠ O A B ) sin ( ∠ A O B ) \displaystyle{ r = \frac{ A B \cdot \sin \left( \angle O A B \right) }{ \sin \left( \angle A O B \right) } } r = sin ( ∠ A OB ) A B ⋅ sin ( ∠ O A B )
sin ( ∠ A O B ) = sin ( 2 ∠ A C B ) = 2 sin ( ∠ A C B ) cos ( ∠ A C B ) \displaystyle{ \begin{aligned}\sin \left( \angle A O B \right) & = \sin \left( 2 \angle A C B \right) \\ & = 2 \sin \left( \angle A C B \right) \cos \left( \angle A C B \right)\end{aligned} } sin ( ∠ A OB ) = sin ( 2∠ A CB ) = 2 sin ( ∠ A CB ) cos ( ∠ A CB )
∠ O A B + ∠ A C B = 90 ° ∠ O A B = 90 ° − ∠ A C B sin ( ∠ O A B ) = sin ( 90 ° − ∠ A C B ) = cos ( ∠ A C B ) \displaystyle{ \begin{aligned}\angle O A B + \angle A C B & = 90 \degree \\ \angle O A B & = 90 \degree - \angle A C B \\ \sin \left( \angle O A B \right) & = \sin \left( 90 \degree - \angle A C B \right) \\ & = \cos \left( \angle A C B \right)\end{aligned} } ∠ O A B + ∠ A CB ∠ O A B sin ( ∠ O A B ) = 90° = 90° − ∠ A CB = sin ( 90° − ∠ A CB ) = cos ( ∠ A CB )
r = A B ⋅ sin ( ∠ O A B ) sin ( ∠ A O B ) = A B ⋅ cos ( ∠ A C B ) 2 sin ( ∠ A C B ) cos ( ∠ A C B ) = A B 2 sin ( ∠ A C B ) \displaystyle{ \begin{aligned}r & = \frac{ A B \cdot \sin \left( \angle O A B \right) }{ \sin \left( \angle A O B \right) } \\ & = \frac{ A B \cdot \cos \left( \angle A C B \right) }{ 2 \sin \left( \angle A C B \right) \cos \left( \angle A C B \right) } \\ & = \frac{ A B }{ 2 \sin \left( \angle A C B \right) }\end{aligned} } r = sin ( ∠ A OB ) A B ⋅ sin ( ∠ O A B ) = 2 sin ( ∠ A CB ) cos ( ∠ A CB ) A B ⋅ cos ( ∠ A CB ) = 2 sin ( ∠ A CB ) A B
Luas segitiga ABC bisa dicari dengan trigonometri:
L A B C = 1 2 ⋅ alas ⋅ tinggi = 1 2 A C ( B C ⋅ sin ( ∠ A C B ) ) sin ( ∠ A C B ) = 2 L A B C A C ⋅ B C \displaystyle{ \begin{aligned}L _{ A B C } & = \frac{ 1 }{ 2 } \cdot \text{alas} \cdot \text{tinggi} \\ & = \frac{ 1 }{ 2 } A C \left( B C \cdot \sin \left( \angle A C B \right) \right) \\ \sin \left( \angle A C B \right) & = \frac{ 2 L _{ A B C } }{ A C \cdot B C }\end{aligned} } L A BC sin ( ∠ A CB ) = 2 1 ⋅ alas ⋅ tinggi = 2 1 A C ( BC ⋅ sin ( ∠ A CB ) ) = A C ⋅ BC 2 L A BC
r = A B 2 sin ( ∠ A C B ) = A B 2 ( 2 L A B C A C ⋅ B C ) = A B ⋅ A C ⋅ B C 4 L A B C \displaystyle{ \begin{aligned}r & = \frac{ A B }{ 2 \sin \left( \angle A C B \right) } \\ & = \frac{ A B }{ 2 \left( \displaystyle \frac{ 2 L _{ A B C } }{ A C \cdot B C } \right) } \\ & = \frac{ A B \cdot A C \cdot B C }{ 4 L _{ A B C } }\end{aligned} } r = 2 sin ( ∠ A CB ) A B = 2 ( A C ⋅ BC 2 L A BC ) A B = 4 L A BC A B ⋅ A C ⋅ BC
Hari hari lingkaran luas yang berada di luar segitiga ABc bisa dicari dengan:
r = A B ⋅ A C ⋅ B C 4 L A B C \displaystyle{ r = \frac{ A B \cdot A C \cdot B C }{ 4 L _{ A B C } } } r = 4 L A BC A B ⋅ A C ⋅ BC
Alternatif Penurunan dengan Hukum Cosinus dan Identitas Trigonometri serta Teorema Heron
Berlaku hukum Cosinus pada segitiga ABC dengan sudut ∠ A B C \displaystyle{ \angle A B C } ∠ A BC
A B 2 = A C 2 + B C 2 − 2 ⋅ A C ⋅ B C ⋅ cos ( ∠ A C B ) cos ( ∠ A C B ) = A B 2 − A C 2 − B C 2 − 2 ⋅ A C ⋅ B C = A C 2 + B C 2 − A B 2 2 ⋅ A C ⋅ B C \displaystyle{ \begin{aligned}A B ^{ 2 } & = A C ^{ 2 } + B C ^{ 2 } - 2 \cdot A C \cdot B C \cdot \cos \left( \angle A C B \right) \\ \cos \left( \angle A C B \right) & = \frac{ A B ^{ 2 } - A C ^{ 2 } - B C ^{ 2 } }{ - 2 \cdot A C \cdot B C } \\ & = \frac{ A C ^{ 2 } + B C ^{ 2 } - A B ^{ 2 } }{ 2 \cdot A C \cdot B C }\end{aligned} } A B 2 cos ( ∠ A CB ) = A C 2 + B C 2 − 2 ⋅ A C ⋅ BC ⋅ cos ( ∠ A CB ) = − 2 ⋅ A C ⋅ BC A B 2 − A C 2 − B C 2 = 2 ⋅ A C ⋅ BC A C 2 + B C 2 − A B 2
Untuk mendapatkan nilai sin dapat digunakan identitas trigonometri:
sin 2 ( ∠ A C B ) + cos 2 ( ∠ A C B ) = 1 sin 2 ( ∠ A C B ) = 1 − cos 2 ( ∠ A C B ) \displaystyle{ \begin{aligned}\sin ^{ 2 } \left( \angle A C B \right) + \cos ^{ 2 } \left( \angle A C B \right) & = 1 \\ \sin ^{ 2 } \left( \angle A C B \right) & = 1 - \cos ^{ 2 } \left( \angle A C B \right)\end{aligned} } sin 2 ( ∠ A CB ) + cos 2 ( ∠ A CB ) sin 2 ( ∠ A CB ) = 1 = 1 − cos 2 ( ∠ A CB )
sin 2 ( ∠ A C B ) = 1 − ( A C 2 + B C 2 − A B 2 2 ⋅ A C ⋅ B C ) 2 = 1 − ( A C 2 + B C 2 − A B 2 ) 2 ( 2 ⋅ A C ⋅ B C ) 2 = ( 2 ⋅ A C ⋅ B C ) 2 − ( A C 2 + B C 2 − A B 2 ) 2 ( 2 ⋅ A C ⋅ B C ) 2 = ( 2 ⋅ A C ⋅ B C + ( A C 2 + B C 2 − A B 2 ) ) ( 2 ⋅ A C ⋅ B C − ( A C 2 + B C 2 − A B 2 ) ) ( 2 ⋅ A C ⋅ B C ) 2 = ( 2 ⋅ A C ⋅ B C + A C 2 + B C 2 − A B 2 ) ( 2 ⋅ A C ⋅ B C − A C 2 − B C 2 + A B 2 ) ( 2 ⋅ A C ⋅ B C ) 2 = ( ( A C 2 + B C 2 + 2 ⋅ A C ⋅ B C ) − A B 2 ) ( − ( A C 2 + B C 2 − 2 ⋅ A C ⋅ B C ) + A B 2 ) ( 2 ⋅ A C ⋅ B C ) 2 = ( ( A C + B C ) 2 − A B 2 ) ( − ( A C − B C ) 2 + A B 2 ) ( 2 ⋅ A C ⋅ B C ) 2 = ( ( A C + B C ) 2 − A B 2 ) ( A B 2 − ( A C − B C ) 2 ) ( 2 ⋅ A C ⋅ B C ) 2 = ( A C + B C + A B ) ( A C + B C − A B ) ( A B + ( A C − B C ) ) ( A B − ( A C − B C ) ) ( 2 ⋅ A C ⋅ B C ) 2 = ( A C + B C + A B ) ( A C + B C − A B ) ( A B + A C − B C ) ( A B − A C + B C ) ( 2 ⋅ A C ⋅ B C ) 2 \displaystyle{ \begin{aligned}\sin ^{ 2 } \left( \angle A C B \right) & = 1 - \left( \frac{ A C ^{ 2 } + B C ^{ 2 } - A B ^{ 2 } }{ 2 \cdot A C \cdot B C } \right) ^{ 2 } \\ & = 1 - \frac{ \left( A C ^{ 2 } + B C ^{ 2 } - A B ^{ 2 } \right) ^{ 2 } }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } - \left( A C ^{ 2 } + B C ^{ 2 } - A B ^{ 2 } \right) ^{ 2 } }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ \left( 2 \cdot A C \cdot B C + \left( A C ^{ 2 } + B C ^{ 2 } - A B ^{ 2 } \right) \right) \left( 2 \cdot A C \cdot B C - \left( A C ^{ 2 } + B C ^{ 2 } - A B ^{ 2 } \right) \right) }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ \left( 2 \cdot A C \cdot B C + A C ^{ 2 } + B C ^{ 2 } - A B ^{ 2 } \right) \left( 2 \cdot A C \cdot B C - A C ^{ 2 } - B C ^{ 2 } + A B ^{ 2 } \right) }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ \left( \left( A C ^{ 2 } + B C ^{ 2 } + 2 \cdot A C \cdot B C \right) - A B ^{ 2 } \right) \left( - \left( A C ^{ 2 } + B C ^{ 2 } - 2 \cdot A C \cdot B C \right) + A B ^{ 2 } \right) }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ \left( \left( A C + B C \right) ^{ 2 } - A B ^{ 2 } \right) \left( - \left( A C - B C \right) ^{ 2 } + A B ^{ 2 } \right) }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ \left( \left( A C + B C \right) ^{ 2 } - A B ^{ 2 } \right) \left( A B ^{ 2 } - \left( A C - B C \right) ^{ 2 } \right) }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ \left( A C + B C + A B \right) \left( A C + B C - A B \right) \left( A B + \left( A C - B C \right) \right) \left( A B - \left( A C - B C \right) \right) }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ \left( A C + B C + A B \right) \left( A C + B C - A B \right) \left( A B + A C - B C \right) \left( A B - A C + B C \right) }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } }\end{aligned} } sin 2 ( ∠ A CB ) = 1 − ( 2 ⋅ A C ⋅ BC A C 2 + B C 2 − A B 2 ) 2 = 1 − ( 2 ⋅ A C ⋅ BC ) 2 ( A C 2 + B C 2 − A B 2 ) 2 = ( 2 ⋅ A C ⋅ BC ) 2 ( 2 ⋅ A C ⋅ BC ) 2 − ( A C 2 + B C 2 − A B 2 ) 2 = ( 2 ⋅ A C ⋅ BC ) 2 ( 2 ⋅ A C ⋅ BC + ( A C 2 + B C 2 − A B 2 ) ) ( 2 ⋅ A C ⋅ BC − ( A C 2 + B C 2 − A B 2 ) ) = ( 2 ⋅ A C ⋅ BC ) 2 ( 2 ⋅ A C ⋅ BC + A C 2 + B C 2 − A B 2 ) ( 2 ⋅ A C ⋅ BC − A C 2 − B C 2 + A B 2 ) = ( 2 ⋅ A C ⋅ BC ) 2 ( ( A C 2 + B C 2 + 2 ⋅ A C ⋅ BC ) − A B 2 ) ( − ( A C 2 + B C 2 − 2 ⋅ A C ⋅ BC ) + A B 2 ) = ( 2 ⋅ A C ⋅ BC ) 2 ( ( A C + BC ) 2 − A B 2 ) ( − ( A C − BC ) 2 + A B 2 ) = ( 2 ⋅ A C ⋅ BC ) 2 ( ( A C + BC ) 2 − A B 2 ) ( A B 2 − ( A C − BC ) 2 ) = ( 2 ⋅ A C ⋅ BC ) 2 ( A C + BC + A B ) ( A C + BC − A B ) ( A B + ( A C − BC ) ) ( A B − ( A C − BC ) ) = ( 2 ⋅ A C ⋅ BC ) 2 ( A C + BC + A B ) ( A C + BC − A B ) ( A B + A C − BC ) ( A B − A C + BC )
A C + B C + A B = K = 2 s A C + B C − A B = A C + B C + A B − 2 A B = 2 s − 2 A B = 2 ( s − A B ) A B + A C − B C = A B + A C + B C − 2 B C = 2 s − 2 B C = 2 ( s − B C ) A C + B C − A B = A C + B C + A B − 2 A B = 2 s − 2 A B = 2 ( s − A B ) A B − A C + B C = A B + A C + B C − 2 A C = 2 s − 2 A C = 2 ( s − A C ) \displaystyle{ \begin{aligned}A C + B C + A B & = K = 2 s \\ A C + B C - A B & = A C + B C + A B - 2 A B \\ & = 2 s - 2 A B \\ & = 2 \left( s - A B \right) \\ A B + A C - B C & = A B + A C + B C - 2 B C \\ & = 2 s - 2 B C \\ & = 2 \left( s - B C \right) \\ A C + B C - A B & = A C + B C + A B - 2 A B \\ & = 2 s - 2 A B \\ & = 2 \left( s - A B \right) \\ A B - A C + B C & = A B + A C + B C - 2 A C \\ & = 2 s - 2 A C \\ & = 2 \left( s - A C \right)\end{aligned} } A C + BC + A B A C + BC − A B A B + A C − BC A C + BC − A B A B − A C + BC = K = 2 s = A C + BC + A B − 2 A B = 2 s − 2 A B = 2 ( s − A B ) = A B + A C + BC − 2 BC = 2 s − 2 BC = 2 ( s − BC ) = A C + BC + A B − 2 A B = 2 s − 2 A B = 2 ( s − A B ) = A B + A C + BC − 2 A C = 2 s − 2 A C = 2 ( s − A C )
sin 2 ( ∠ A C B ) = ( A C + B C + A B ) ( A C + B C − A B ) ( A B + A C − B C ) ( A B − A C + B C ) ( 2 ⋅ A C ⋅ B C ) 2 = ( 2 s ) ( 2 ( s − A B ) ) ( 2 ( s − B C ) ) ( 2 ( s − A C ) ) ( 2 ⋅ A C ⋅ B C ) 2 = 16 s ( s − A B ) ( s − B C ) ( s − A C ) 4 ⋅ A C 2 ⋅ B C 2 = 4 s ( s − A B ) ( s − B C ) ( s − A C ) A C 2 ⋅ B C 2 sin ( ∠ A C B ) = 4 s ( s − A B ) ( s − B C ) ( s − A C ) A C 2 ⋅ B C 2 = 2 s ( s − A B ) ( s − B C ) ( s − A C ) A C ⋅ B C = 2 s ( s − A B ) ( s − B C ) ( s − A C ) A C ⋅ B C = 2 L A B C A C ⋅ B C \displaystyle{ \begin{aligned}\sin ^{ 2 } \left( \angle A C B \right) & = \frac{ \left( A C + B C + A B \right) \left( A C + B C - A B \right) \left( A B + A C - B C \right) \left( A B - A C + B C \right) }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ \left( 2 s \right) \left( 2 \left( s - A B \right) \right) \left( 2 \left( s - B C \right) \right) \left( 2 \left( s - A C \right) \right) }{ \left( 2 \cdot A C \cdot B C \right) ^{ 2 } } \\ & = \frac{ 16 s \left( s - A B \right) \left( s - B C \right) \left( s - A C \right) }{ 4 \cdot A C ^{ 2 } \cdot B C ^{ 2 } } \\ & = \frac{ 4 s \left( s - A B \right) \left( s - B C \right) \left( s - A C \right) }{ A C ^{ 2 } \cdot B C ^{ 2 } } \\ \sin \left( \angle A C B \right) & = \sqrt{ \frac{ 4 s \left( s - A B \right) \left( s - B C \right) \left( s - A C \right) }{ A C ^{ 2 } \cdot B C ^{ 2 } } } \\ & = \frac{ 2 \sqrt{ s \left( s - A B \right) \left( s - B C \right) \left( s - A C \right) } }{ A C \cdot B C } \\ & = \frac{ 2 \sqrt{ s \left( s - A B \right) \left( s - B C \right) \left( s - A C \right) } }{ A C \cdot B C } \\ & = \frac{ 2 L _{ A B C } }{ A C \cdot B C }\end{aligned} } sin 2 ( ∠ A CB ) sin ( ∠ A CB ) = ( 2 ⋅ A C ⋅ BC ) 2 ( A C + BC + A B ) ( A C + BC − A B ) ( A B + A C − BC ) ( A B − A C + BC ) = ( 2 ⋅ A C ⋅ BC ) 2 ( 2 s ) ( 2 ( s − A B ) ) ( 2 ( s − BC ) ) ( 2 ( s − A C ) ) = 4 ⋅ A C 2 ⋅ B C 2 16 s ( s − A B ) ( s − BC ) ( s − A C ) = A C 2 ⋅ B C 2 4 s ( s − A B ) ( s − BC ) ( s − A C ) = A C 2 ⋅ B C 2 4 s ( s − A B ) ( s − BC ) ( s − A C ) = A C ⋅ BC 2 s ( s − A B ) ( s − BC ) ( s − A C ) = A C ⋅ BC 2 s ( s − A B ) ( s − BC ) ( s − A C ) = A C ⋅ BC 2 L A BC
r = A B 2 sin ( ∠ A C B ) = A B 2 ( 2 L A B C A C ⋅ B C ) = A B ⋅ A C ⋅ B C 4 L A B C \displaystyle{ \begin{aligned}r & = \frac{ A B }{ 2 \sin \left( \angle A C B \right) } \\ & = \frac{ A B }{ 2 \left( \displaystyle \frac{ 2 L _{ A B C } }{ A C \cdot B C } \right) } \\ & = \frac{ A B \cdot A C \cdot B C }{ 4 L _{ A B C } }\end{aligned} } r = 2 sin ( ∠ A CB ) A B = 2 ( A C ⋅ BC 2 L A BC ) A B = 4 L A BC A B ⋅ A C ⋅ BC
