In \( \triangle \mathrm{MNO}, n=71 \) inches, \( m \angle \mathrm{~N}=136^{\circ} \) and \( m \) \( \angle \mathrm{O}=42^{\circ} \). Find the length of \( m \), to the nearest 10 th of an inch.

To find the length of side \( m \) in triangle \( \triangle MNO \), we can use the Law of Sines, which states: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] In this case, we have: - \( n = 71 \) inches (opposite angle \( \angle M \)) - \( m \angle N = 136^{\circ} \) - \( m \angle O = 42^{\circ} \) First, we need to calculate \( \angle M \): \[ m \angle M = 180^{\circ} - m \angle N - m \angle O = 180^{\circ} - 136^{\circ} - 42^{\circ} = 2^{\circ} \] Now, we can use the Law of Sines to find side \( m \): \[ \frac{m}{\sin(136^{\circ})} = \frac{71}{\sin(2^{\circ})} \] Solving for \( m \): \[ m = \frac{71 \cdot \sin(136^{\circ})}{\sin(2^{\circ})} \] Using a calculator, we find: - \( \sin(136^{\circ}) \approx 0.8192 \) - \( \sin(2^{\circ}) \approx 0.0349 \) Now plug in the values: \[ m \approx \frac{71 \cdot 0.8192}{0.0349} \approx \frac{58.8944}{0.0349} \approx 1684.62 \] Finally, rounding to the nearest tenth of an inch, we find: \[ m \approx 1684.6 \text{ inches} \]