As shown in the figure, $$\overline{\mathrm{AB}} = 3, \quad \overline{\mathrm{BC}} = \sqrt{13}, \quad \overline{\mathrm{AD}} \times \overline{\mathrm{CD}} = 9, \quad \angle\mathrm{BAC} = \frac{\pi}{3}$$ For quadrilateral ABCD, let $S_1$ denote the area of triangle ABC, $S_2$ denote the area of triangle ACD, and $R$ denote the circumradius of triangle ACD. If $S_2 = \frac{5}{6}S_1$, find the value of $\frac{R}{\sin(\angle\mathrm{ADC})}$. [4 points] (1) $\frac{54}{25}$ (2) $\frac{117}{50}$ (3) $\frac{63}{25}$ (4) $\frac{27}{10}$ (5) $\frac{72}{25}$
As shown in the figure,
$$\overline{\mathrm{AB}} = 3, \quad \overline{\mathrm{BC}} = \sqrt{13}, \quad \overline{\mathrm{AD}} \times \overline{\mathrm{CD}} = 9, \quad \angle\mathrm{BAC} = \frac{\pi}{3}$$
For quadrilateral ABCD, let $S_1$ denote the area of triangle ABC, $S_2$ denote the area of triangle ACD, and $R$ denote the circumradius of triangle ACD.\\
If $S_2 = \frac{5}{6}S_1$, find the value of $\frac{R}{\sin(\angle\mathrm{ADC})}$. [4 points]\\
(1) $\frac{54}{25}$\\
(2) $\frac{117}{50}$\\
(3) $\frac{63}{25}$\\
(4) $\frac{27}{10}$\\
(5) $\frac{72}{25}$