Jawaban:
D
Penjelasan dengan langkah-langkah:
Pertama-tama dihitung panjang BD dg teorema Pythagoras
[tex]BD= \sqrt{ {AB}^{2} + {AD}^{2} } = \sqrt{ {12}^{2} + {5}^{2} } = \sqrt{144 + 25} = \sqrt{169} = 13 \: cm[/tex]
Kemudian menggunakan aturan sinus
[tex] \frac{BD}{sin \: C} = \frac{BC}{sin \: \angle BDC} \\ \frac{13}{sin \: 90^\circ} = \frac{BC}{sin \: 60^\circ} \\ \frac{13}{1} = \frac{BC}{ \frac{1}{2} \sqrt{3} } \\ BC = 13 \times \frac{1}{2} \sqrt{3} = \frac{13 \sqrt{3} }{2} cm[/tex]
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Jawaban:
D
Penjelasan dengan langkah-langkah:
Pertama-tama dihitung panjang BD dg teorema Pythagoras
[tex]BD= \sqrt{ {AB}^{2} + {AD}^{2} } = \sqrt{ {12}^{2} + {5}^{2} } = \sqrt{144 + 25} = \sqrt{169} = 13 \: cm[/tex]
Kemudian menggunakan aturan sinus
[tex] \frac{BD}{sin \: C} = \frac{BC}{sin \: \angle BDC} \\ \frac{13}{sin \: 90^\circ} = \frac{BC}{sin \: 60^\circ} \\ \frac{13}{1} = \frac{BC}{ \frac{1}{2} \sqrt{3} } \\ BC = 13 \times \frac{1}{2} \sqrt{3} = \frac{13 \sqrt{3} }{2} cm[/tex]