Soal yang Akan Dibahas
Diketahi $\Delta ABC$ dan $ \alpha , \, \beta , \, \gamma $
adalah sudut di A, B, dan C. Jika diketahui
$ \sin \beta = \frac{1}{3} $ dan $ \sin \gamma = \frac{1}{2}$ ,
maka $ \frac{BC}{AC} \, $ adalah ....
A). $\frac{1}{2} (\sqrt{3} - 2\sqrt{2}) \, $
B). $\frac{1}{2} (\sqrt{3} - \sqrt{2}) \, $
C). $\frac{1}{2} (\sqrt{3} + 2\sqrt{2}) \, $
D). $(\sqrt{3} + 2\sqrt{2}) \, $
E). $(\sqrt{3} - \sqrt{2}) $
A). $\frac{1}{2} (\sqrt{3} - 2\sqrt{2}) \, $
B). $\frac{1}{2} (\sqrt{3} - \sqrt{2}) \, $
C). $\frac{1}{2} (\sqrt{3} + 2\sqrt{2}) \, $
D). $(\sqrt{3} + 2\sqrt{2}) \, $
E). $(\sqrt{3} - \sqrt{2}) $
$\spadesuit $ Konsep Dasar
*). Identitas Trigonometri
$ \sin ^2 A + \cos ^2 A = 1 \rightarrow \cos A = \sqrt{1 - \sin ^2 A } $
*). Nilai $ \sin (180^\circ - x ) = \sin x $
*). Aturan sinus pada segitiga ABC :
$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $
*). Rumus Jumlah Sudut Trigonometri
$ \sin (A + B ) = \sin A \cos B + \cos A \sin B $
*). Identitas Trigonometri
$ \sin ^2 A + \cos ^2 A = 1 \rightarrow \cos A = \sqrt{1 - \sin ^2 A } $
*). Nilai $ \sin (180^\circ - x ) = \sin x $
*). Aturan sinus pada segitiga ABC :
$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $
*). Rumus Jumlah Sudut Trigonometri
$ \sin (A + B ) = \sin A \cos B + \cos A \sin B $
$\clubsuit $ Pembahasan
*). Ilustrasi gambar.
*). Menentukan nilai $ \cos \beta \, $ dan $ \cos \gamma $ :
$\begin{align} \sin \beta = \frac{1}{3} \rightarrow \cos \beta & = \sqrt{ 1 - \sin ^2 \beta } \\ & = \sqrt{ 1 - (\frac{1}{3})^2 } \\ & = \sqrt{ 1 - \frac{1}{9} } \\ & = \sqrt{ \frac{8}{9} } \\ & = \frac{2\sqrt{ 2}}{3} = \frac{2}{3}\sqrt{2} \\ \sin \gamma = \frac{1}{2} \rightarrow \cos \gamma & = \sqrt{ 1 - \sin ^2 \gamma } \\ & = \sqrt{ 1 - (\frac{1}{2})^2 } \\ & = \sqrt{ 1 - \frac{1}{4} } \\ & = \sqrt{ \frac{3}{9} } \\ & = \frac{\sqrt{ 3}}{2} = \frac{1}{2}\sqrt{3} \end{align} $
*). Menentukan nilai $ \sin \alpha $ :
$\begin{align} \alpha + \beta + \gamma & = 180^\circ \, \, \, \, \text{(jumlah sudut segitiga)} \\ \alpha & = 180^\circ - ( \beta + \gamma ) \\ \sin \alpha & = \sin [180^\circ - ( \beta + \gamma )] \\ \sin \alpha & = \sin ( \beta + \gamma ) \\ & = \sin \beta \cos \gamma + \cos \beta \sin \gamma \\ & = \frac{1}{3} . \frac{1}{2}\sqrt{3} + \frac{2}{3}\sqrt{2} . \frac{1}{2} \\ & = \frac{1}{6}\sqrt{3} + \frac{1}{3}\sqrt{2} \end{align} $
*). Menentukan nilai $ \frac{BC}{AC} $ dengan aturan sinus :
$\begin{align} \frac{BC}{\sin A } & = \frac{AC}{\sin B } \\ \frac{BC}{\sin \alpha } & = \frac{AC}{\sin \beta } \\ \frac{BC}{AC } & = \frac{\sin \alpha}{\sin \beta } \\ & = \frac{\frac{1}{6}\sqrt{3} + \frac{1}{3}\sqrt{2}}{ \frac{1}{3} } \\ & = 3 ( \frac{1}{6}\sqrt{3} + \frac{1}{3}\sqrt{2} ) \\ & = \frac{1}{2}\sqrt{3} + \sqrt{2} \\ & = \frac{1}{2}(\sqrt{3} + 2 \sqrt{2} ) \end{align} $
Jadi, nilai $ \frac{BC}{AC } = \frac{1}{2}(\sqrt{3} + 2 \sqrt{2} ) . \, \heartsuit $
*). Ilustrasi gambar.
*). Menentukan nilai $ \cos \beta \, $ dan $ \cos \gamma $ :
$\begin{align} \sin \beta = \frac{1}{3} \rightarrow \cos \beta & = \sqrt{ 1 - \sin ^2 \beta } \\ & = \sqrt{ 1 - (\frac{1}{3})^2 } \\ & = \sqrt{ 1 - \frac{1}{9} } \\ & = \sqrt{ \frac{8}{9} } \\ & = \frac{2\sqrt{ 2}}{3} = \frac{2}{3}\sqrt{2} \\ \sin \gamma = \frac{1}{2} \rightarrow \cos \gamma & = \sqrt{ 1 - \sin ^2 \gamma } \\ & = \sqrt{ 1 - (\frac{1}{2})^2 } \\ & = \sqrt{ 1 - \frac{1}{4} } \\ & = \sqrt{ \frac{3}{9} } \\ & = \frac{\sqrt{ 3}}{2} = \frac{1}{2}\sqrt{3} \end{align} $
*). Menentukan nilai $ \sin \alpha $ :
$\begin{align} \alpha + \beta + \gamma & = 180^\circ \, \, \, \, \text{(jumlah sudut segitiga)} \\ \alpha & = 180^\circ - ( \beta + \gamma ) \\ \sin \alpha & = \sin [180^\circ - ( \beta + \gamma )] \\ \sin \alpha & = \sin ( \beta + \gamma ) \\ & = \sin \beta \cos \gamma + \cos \beta \sin \gamma \\ & = \frac{1}{3} . \frac{1}{2}\sqrt{3} + \frac{2}{3}\sqrt{2} . \frac{1}{2} \\ & = \frac{1}{6}\sqrt{3} + \frac{1}{3}\sqrt{2} \end{align} $
*). Menentukan nilai $ \frac{BC}{AC} $ dengan aturan sinus :
$\begin{align} \frac{BC}{\sin A } & = \frac{AC}{\sin B } \\ \frac{BC}{\sin \alpha } & = \frac{AC}{\sin \beta } \\ \frac{BC}{AC } & = \frac{\sin \alpha}{\sin \beta } \\ & = \frac{\frac{1}{6}\sqrt{3} + \frac{1}{3}\sqrt{2}}{ \frac{1}{3} } \\ & = 3 ( \frac{1}{6}\sqrt{3} + \frac{1}{3}\sqrt{2} ) \\ & = \frac{1}{2}\sqrt{3} + \sqrt{2} \\ & = \frac{1}{2}(\sqrt{3} + 2 \sqrt{2} ) \end{align} $
Jadi, nilai $ \frac{BC}{AC } = \frac{1}{2}(\sqrt{3} + 2 \sqrt{2} ) . \, \heartsuit $
Tidak ada komentar:
Posting Komentar
Catatan: Hanya anggota dari blog ini yang dapat mengirim komentar.