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SMK Teknologi dan Rekayasa

Solusi Paket 59 UN 2011 Matematika SMK Teknologi (27)


Diketahui\ \ sin\ A=\frac{5}{13}\ \ dan\ \ tan\ B=-\frac{4}{3}.\ Jika\ A\ di\ kuadran\ I\ dan\ B\ di\ kuadran\ II\ maka\\nilai\ sin\ (A-B)\ adalah\ ....\\A.\quad -\frac{63}{65}\quad \quad \quad B.\quad -\frac{33}{65}\quad \quad \quad C.\quad -\frac{16}{65}\quad \quad \quad D.\quad \frac{33}{65}\quad \quad \quad E.\quad \frac{56}{65}
Jawaban: A
sin\ (A-B)=sin\ A\ cos\ B-cos\ A\ sin\ B
Dengan\ menggunakan\ dalil\ Pythagoras,\ diperoleh\ tripel\ bilangan\ Pythagoras:\\\begin{matrix} & & & & & & & & & \end{matrix}5,\ \mathbf{12},\ 13\begin{matrix} & & & & & & & & & \end{matrix}3,\ 4,\ \mathbf{5}
sin\ A=\frac{5}{13}\begin{matrix} & \end{matrix}\to \begin{matrix} & \end{matrix}cos\ A=\frac{12}{13}\\\\(nilai\ cos\ A\ positif\ karena\ A\ sudut\ di\ kuadran\ I)
tan\ B=-\frac{4}{3}\begin{matrix} & \end{matrix}\to \begin{matrix} & \end{matrix}sin\ B=\frac{4}{5}\qquad dan\qquad cos\ B=-\frac{3}{5}\\\\(nilai\ sin\ B\ positif\ dan\ nilai\ cos\ B\ negatif\ karena\ B\ sudut\ di\ kuadran\ II)
\begin{array}{rcl}sin\ (A-B)&=&\frac{5}{13}\ \times \ \big(-\frac{3}{5}\big)\ \ -\ \ \frac{12}{13}\ \times \ \frac{4}{5}\\\\&=&-\frac{15}{65}\ \ -\ \ \frac{48}{65}\\\\&=&-\frac{63}{65}\end{array}

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