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SMK Teknologi dan Rekayasa

Soal 85 Perbandingan Trigonometri


Diketahui\ segitiga\ ABC,\ \angle A=30^{\circ},\ \angle B=90^{\circ},\ dan\ panjang\ AB=6\ cm.\\Panjang\ AC+BC=\ ....\\a.\ \ 3\sqrt{3}\ \ \ \ \ \ \ \ \ \ b.\ \ 4\sqrt{3}\ \ \ \ \ \ \ \ \ \ c.\ \ 5\sqrt{3}\ \ \ \ \ \ \ \ \ \ d.\ \ 6\sqrt{3}\ \ \ \ \ \ \ \ \ \ e.\ \ 7\sqrt{3}
Jawaban: d


\begin{array}{rcl}\frac{AB}{AC}&=&cos\ 30^{\circ}\\\\\frac{6}{AC}&=&\frac{1}{2}\sqrt{3}\\\\AC&=&\frac{6}{\frac{1}{2}\sqrt{3}}\ \ \ \ \times \ \ \frac{2\sqrt{3}}{2\sqrt{3}}\\\\&=&\frac{6\times 2\times \sqrt{3}}{\frac{1}{2}\times 2\times \sqrt{3}\times \sqrt{3}}\\\\&=&\frac{12\sqrt{3}}{3}\\\\&=&4\sqrt{3}\end{array}
Berdasarkan\ dalil\ Pythagoras:
\begin{array}{rcl}BC^{2}&=&AC^{2}-AB^{2}\\\\&=&(4\sqrt{3})^{2}-6^{2}\\\\&=&16.3-36\\\\&=&48-36\\\\&=&12\\\\BC&=&\sqrt{12}\\\\&=&\sqrt{4.3}\\\\&=&\sqrt{4}\times \sqrt{3}\\\\&=&2\sqrt{3}\end{array}
AC+BC=4\sqrt{3}+2\sqrt{3}=6\sqrt{3}

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One thought on “Soal 85 Perbandingan Trigonometri

  1. Makasih ya soal-soal dan pembahasannya. Sangat membantu🙂

    Posted by Silent Girl | 20 Mei 2015, 12:04 AM

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