SNBT Zone : Balok
Table of Contents

Tipe:
No.
Balok ABCD.EFGH alasnya persegi dengan AB = 4 cm, AE = 8 cm. Jika θ adalah sudut antara bidang ACH dan bidang ABCD, maka sin θ = ....\dfrac45\sqrt2 \dfrac23\sqrt2 \dfrac12\sqrt2
\dfrac47\sqrt2 \dfrac13
ALTERNATIF PENYELESAIAN
θ = ∠ DOH
BD=4\sqrt2
\begin{aligned}
DO&=\dfrac12BD\\[3.5pt]
&=\dfrac12\left(4\sqrt2\right)\\[3.5pt]
&=2\sqrt2
\end{aligned}
\begin{aligned}
HO&=\sqrt{HD^2+DO^2}\\
&=\sqrt{8^2+\left(2\sqrt2\right)^2}\\
&=\sqrt{64+8}\\
&=\sqrt{72}\\
&=6\sqrt2
\end{aligned}
\begin{aligned}
\sin\theta&=\dfrac{HD}{HO}\\[3.5pt]
&=\dfrac8{6\sqrt2}\\[3.5pt]
&=\dfrac4{3\sqrt2}&\times\dfrac{\sqrt2}{\sqrt2}\\[3.5pt]
&=\dfrac46\sqrt2\\
&=\boxed{\boxed{\dfrac23\sqrt2}}
\end{aligned}
Jadi, \sin\theta =\dfrac23\sqrt2 .
JAWAB: B
JAWAB: B
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