In a triangle $\mathrm{ABC}$, with usual notations, if $\mathrm{m} \angle \mathrm{A}=60^{\circ},…
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
$\frac{\sqrt{3}}{2}$
$\frac{2}{\sqrt{3}}$
$2 \sqrt{3}$
$\frac{1}{2 \sqrt{3}}$
Solution
By sine rule, we get
$\begin{aligned}
& \frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}} \\
& \Rightarrow \frac{\sin 60^{\circ}}{6}=\frac{x}{8} \\
& \Rightarrow x=\frac{\sqrt{3}}{2} \times \frac{8}{6} \\
& \Rightarrow x=\frac{2}{\sqrt{3}}
\end{aligned}$