In \( \triangle \mathrm{EFG}, g=9.9 \) inches, \( f=6.1 \) inches and \( \angle \mathrm{F}=32^{\circ} \). Find all possible values of \( \angle \mathrm{G} \), to the nearest 1oth of a degree.

To solve for the possible values of \( \angle G \) in triangle \( \mathrm{EFG} \) given \( g=9.9 \) inches, \( f=6.1 \) inches, and \( \angle F=32^\circ \), we can use the Law of Sines. According to this law: \[ \frac{f}{\sin F} = \frac{g}{\sin G} \] Plugging in the values we have: \[ \frac{6.1}{\sin(32^\circ)} = \frac{9.9}{\sin G} \] Calculating \( \sin(32^\circ) \), we find it approximately \( 0.5299 \). Thus, setting up the proportion: \[ \frac{6.1}{0.5299} = \frac{9.9}{\sin G} \] This simplifies to: \[ \sin G = \frac{9.9 \cdot 0.5299}{6.1} \] Calculating \( \sin G \): \[ \sin G \approx \frac{5.2431}{6.1} \approx 0.8582 \] To find \( \angle G \), we take the inverse sine: \[ \angle G \approx \sin^{-1}(0.8582) \approx 60.9^\circ \] However, since sine is positive in both the first and second quadrants, we also consider the supplementary angle: \[ \angle G' = 180^\circ - 60.9^\circ \approx 119.1^\circ \] Thus, the possible values of \( \angle G \) are approximately \( 60.9^\circ \) and \( 119.1^\circ \), rounding to the nearest tenth of a degree.