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SMK Teknologi dan Rekayasa

Solusi Paket 59 UN 2011 Matematika SMK Teknologi (16)


Sebuah\ pohon\ tumbang\ tersandar\ pada\ pagar\ membentuk\ sudut\ 60^{\circ}\ dengan\ tanah.\\Jika\ tinggi\ pagar\ 4\ m,\ maka\ jarak\ pangkal\ pohon\ dengan\ pagar\ (x)\ adalah\ ....

A.\ \ 2\sqrt{3}\ \ \ \ \ \ \ \ \ \ B.\ \ \frac{4}{3}\sqrt{3}\ \ \ \ \ \ \ \ \ \ C.\ \ \frac{8}{3}\sqrt{3}\ \ \ \ \ \ \ \ \ \ D.\ \ 4\sqrt{3}\ \ \ \ \ \ \ \ \ \ E.\ \ 8\sqrt{3}
Jawaban: B
\begin{array}{rcl}\frac{4}{x}&=&tan\ 60^{\circ}\\\\\frac{4}{x}&=&\sqrt{3}\\\\\frac{x}{4}&=&\frac{1}{\sqrt{3}}\qquad \qquad ............\ \ kalikan\ dengan\ \ 4\\\\x&=&\frac{4}{\sqrt{3}}\qquad \qquad ............\ \ kalikan\ dengan\ \ \frac{\sqrt{3}}{\sqrt{3}}\\\\x&=&\frac{4\sqrt{3}}{3}\\\\x&=&\frac{4}{3}\sqrt{3}\end{array}

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