7. In \( \triangle \mathrm{EFG}, f=3.4 \mathrm{~cm}, m \angle \mathrm{~F}=81^{\circ} \) and \( m \angle \mathrm{G}=62^{\circ} \). Find the length of \( e \), to the nearest 10 th of a centimeter.

To solve for the length of side \( e \) in triangle \( \triangle EFG \), we can use the Law of Sines, which states: \[ \frac{e}{\sin E} = \frac{f}{\sin F} \] First, we need to determine \( m \angle E \). We can find \( m \angle E \) using the fact that the sum of the angles in a triangle is \( 180^{\circ} \): \[ m \angle E = 180^{\circ} - m \angle F - m \angle G = 180^{\circ} - 81^{\circ} - 62^{\circ} = 37^{\circ} \] Next, we can plug in the values into the Law of Sines: \[ \frac{e}{\sin 37^{\circ}} = \frac{3.4 \, \text{cm}}{\sin 81^{\circ}} \] Now, we can calculate \( \sin 37^{\circ} \) and \( \sin 81^{\circ} \). Using a calculator, we find: - \( \sin 37^{\circ} \approx 0.6018 \) - \( \sin 81^{\circ} \approx 0.9877 \) Now substitute these values: \[ \frac{e}{0.6018} = \frac{3.4}{0.9877} \] To solve for \( e \): \[ e = 0.6018 \cdot \frac{3.4}{0.9877} \approx 0.6018 \cdot 3.44 \approx 2.07 \, \text{cm} \] Rounding to the nearest tenth of a centimeter, the length of \( e \) is approximately: \[ \boxed{2.1 \, \text{cm}} \]