Kaa plis jawabb :( mtk kelas 10 terimakasiii
Penjelasan dengan langkah-langkah:
[tex]sin \: a = \frac{bc}{ab} \\ sin \: 60 = \frac{3 \sqrt{6} }{ab} \\ \frac{1}{2} \sqrt{3} = \frac{3 \sqrt{6} }{ab} \\ \sqrt{3}ab = 6 \sqrt{6} \\ ab = \frac{6 \sqrt{3 \times 2} }{ \sqrt{3} } \\ ab = 6 \sqrt{2} \: cm [/tex]
[tex]cos \: 45 = \frac{bd}{ab} \\ \frac{1}{2} \sqrt{2} = \frac{bd}{6 \sqrt{2} } \\ \frac{6 \sqrt{2}( \sqrt{2)} }{2} = bd \\ bd = 3 \times 2 \\ bd = 6 \: cm [/tex]
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