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SMUP 2010

Solusi SMUP 2010 Kode 041 – Nomor 5


Dalam\ segitiga\ ABC,\ jika\ panjang\ AB=6,\ \angle A=30^\circ,\ \angle C=120^\circ,\\maka\ luas\ segitiga\ ABC\ sama\ dengan\ ....\ satuan\ luas.\\A.\quad 2\sqrt{3}\\B.\quad 3\sqrt{3}\\C.\quad 6\sqrt{3}\\D.\quad 9\\E.\quad 18
Jawaban: B

\angle B=180^\circ -150^\circ =30^\circ
Mencari panjang AC:

\begin{array} {rcl} \frac{AC}{sinB}&=&\frac{AB}{sinC}\\\\\frac{AC}{sin30^\circ }&=&\frac{6}{sin120^\circ }\\\\\frac{AC}{\frac{1}{2}}&=&\frac{6}{\frac{1}{2}\sqrt{3}}\\\\AC&=&\frac{6}{\sqrt{3}}\\\\AC&=&2\sqrt{3}\end{array}
Luas daerah segitiga ABC:

\begin{array} {rcl} L&=&\frac{1}{2}\times AC\times AB\times sin\ 30^\circ \\\\&=&\frac{1}{2}\times 2\sqrt{3}\times 6\times \frac{1}{2}\\\\&=&3\sqrt{3}\end{array}

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