In $\triangle \mathrm{ABC}$, with usual notations, if $\mathrm{b}=3$, $\mathrm{c}=8, \mathrm{~m} \angle…
In $\triangle \mathrm{ABC}$, with usual notations, if $\mathrm{b}=3$, $\mathrm{c}=8, \mathrm{~m} \angle \mathrm{~A}=60^{\circ}$, then the circumradius of the triangle is $\qquad$ units.
$\frac{7}{3}$
$\frac{7 \sqrt{2}}{3}$
$\frac{7}{\sqrt{3}}$
$\frac{7 \sqrt{3}}{2}$
Solution
By cosine rule, we get
$\begin{aligned}
a^2 & =b^2+c^2-2 b c \cos A \\
& =9+64-48 \cos 60^{\circ} \\
& =73-48 \times \frac{1}{2} \\
& =73-24 \\
& =49 \\
a & =7
\end{aligned}$
$\begin{aligned}
\text {By sine rule, circumradius } & =\frac{a}{\sin \mathrm{~A}} \\
& =\frac{7}{2 \sin 60^{\circ}} \\
& =\frac{7}{2 \times \frac{\sqrt{3}}{2}} \\
& =\frac{7}{\sqrt{3}}
\end{aligned}$