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In a triangle $\mathrm{ABC}$, with usual notations, if $\mathrm{m} \angle \mathrm{A}=60^{\circ}, \mathrm{b}=8, \mathrm{a}=6$ and $\mathrm{B}=\sin ^{-1} x$, then $x$ has the value
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Verified Answer
The correct answer is:
$\frac{2}{\sqrt{3}}$
By sine rule, we get
$$
\begin{aligned}
& \frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}} \\
& \Rightarrow \frac{\sin 60^{\circ}}{6}=\frac{x}{8} \\
& \Rightarrow x=\frac{\sqrt{3}}{2} \times \frac{8}{6} \\
& \Rightarrow x=\frac{2}{\sqrt{3}}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}} \\
& \Rightarrow \frac{\sin 60^{\circ}}{6}=\frac{x}{8} \\
& \Rightarrow x=\frac{\sqrt{3}}{2} \times \frac{8}{6} \\
& \Rightarrow x=\frac{2}{\sqrt{3}}
\end{aligned}
$$
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