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SMK Teknologi dan Rekayasa

Soal 89 Koordinat Kutub – Koordinat Kartesius


Koordinat\ kutub\ titik\ P(4,\ 150^{\circ}),\ maka\ koordinat\ kartesiusnya\\adalah\ ....\\a.\ \ (2\sqrt{3},\ 2)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b.\ \ (-2\sqrt{3},\ 2)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c.\ \ (2\sqrt{3},\ -2)\\d.\ \ (-2\sqrt{3},\ -2)\ \ \ \ \ \ \ \ \ \ \ e.\ \ (2,\ 2\sqrt{3})
Jawaban: b

\begin{array}{rcl}Koordinat\ kutub&\rightarrow &Koordinat\ kartesius\\(r,\ \alpha )&\rightarrow &(x,\ y)\end{array}
r=4,\ \ \alpha =150^{\circ }\\(Karena\ \alpha \ sudut\ di\ kuadran\ II,\ maka\ x\ negatif\ dan\ y\ positif)
\begin{array}{rcl}x&=&r\ cos\ \alpha \\&=&4\times cos\ 150^{\circ }\\&=&4\times cos\ (180-30)^{\circ }\\&=&4\times (-cos\ 30^{\circ })\\&=&4\times (-\frac{1}{2}\sqrt{3})\\&=&-2\sqrt{3}\end{array}
\begin{array}{rcl}y&=&r\ sin\ \alpha \\&=&4\times sin\ 150^{\circ }\\&=&4\times \ sin\ 30^{\circ }\\&=&4\times \frac{1}{2}\\&=&2\end{array}
Koordinat\ kartesiusnya\ adalah\ (-2\sqrt{3},\ 2)

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