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SMK Teknologi dan Rekayasa

Soal 95 Rumus Trigonometri Jumlah-Selisih Sudut


Diketahui\ \ sin\ A=\frac{3}{5}\ \ \ dan\ \ cos\ B=\frac{3}{5},\ \ A\ sudut\ lancip\ dan\\\\B\ sudut\ tumpul.\ Nilai\ cos\ (A-B)=\ ....
a.\ -\frac{24}{25}\ \ \ \ \ \ \ \ \ \ b.\ -\frac{14}{25}\ \ \ \ \ \ \ \ \ \ c.\ \ 0\ \ \ \ \ \ \ \ \ \ d.\ \ \frac{14}{25}\ \ \ \ \ \ \ \ \ \ e.\ \ \frac{24}{25}
Jawaban: c
Dengan\ menggunakan\ dalil\ Pythagoras,\ kita\ bisa\ mengingat\\tripel\ bilangan\ Pythagoras:\\\begin{matrix} & & & & & & & & & \end{matrix}3,\ \mathbf{4},\ 5
sin\ A=\frac{3}{5}\begin{matrix} & \end{matrix}\to \begin{matrix} & \end{matrix}cos\ A=\frac{4}{5}
cos\ B=-\frac{3}{5}\begin{matrix} & \end{matrix}\to \begin{matrix} & \end{matrix}sin\ B=\frac{4}{5}
\begin{array}{rcl}cos\ (A-B)&=&cos\ A\ cos\ B-sin\ A\ sin\ B\\\\&=&\frac{4}{5}\ \times \ \left (-\frac{3}{5} \right )\ \ +\ \ \frac{3}{5}\ \times \ \frac{4}{5}\\\\&=&-\frac{12}{25}\ \ +\ \ \frac{12}{25}\\\\&=&0\end{array}
\mathbf{Keterangan}:\\Nilai\ cos\ A\ bertanda\ positif,\ karena\ A\ sudut\ lancip\\(sudut\ di\ kuadran\ I)\\Nilai\ sin\ B\ juga\ bertanda\ positif,\ karena\ B\ sudut\ tumpul\\(sudut\ di\ kuadran\ II)

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