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SMK Teknologi dan Rekayasa

Solusi Paket 59 UN 2011 Matematika SMK Teknologi (26)


Koordinat\ kartesius\ dari\ titik\ (6,\ 300^{\circ})\ adalah\ ....\\A.\quad (-3\sqrt{3},\ 3)\quad \quad B.\quad (3,\ 3\sqrt{3})\quad \quad C.\quad (3,\ -3\sqrt{3})\quad \quad D.\quad (3\sqrt{3},\ -3)\quad \quad E.\quad (-3,\ -3\sqrt{3})
Jawaban: C
\begin{array}{rcl}Koordinat\ kutub&\rightarrow &Koordinat\ kartesius\\(r,\ \alpha )&\rightarrow &(x,\ y)\end{array}
r=6,\ \ \alpha =300^{\circ }\\(Karena\ \alpha \ sudut\ di\ kuadran\ IV,\ maka\ x\ positif\ dan\ y\ negatif)
\begin{array}{rcl}x&=&r\ cos\ \alpha \\&=&4\times cos\ 300^{\circ }\\&=&6\times cos\ (360-60)^{\circ }\\&=&6\times (+cos\ 60^{\circ })\\&=&6\times \frac{1}{2}\\&=&3\end{array}
\begin{array}{rcl}y&=&r\ sin\ \alpha \\&=&6\times sin\ 300^{\circ }\\&=&6\times \ (-sin\ 60^{\circ })\\&=&6\times \big(-\frac{1}{2}\sqrt{3}\big)\\&=&-3\sqrt{3}\end{array}
Koordinat\ kartesiusnya\ adalah\ (3,\ -3\sqrt{3})

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