3. In \( \triangle \mathrm{NOP}, n=3.1 \mathrm{~cm}, o=5.9 \mathrm{~cm} \) and \( \angle \mathrm{P}=171^{\circ} \). Find the length of \( p \), to the nearest 10 th of a centimeter.

To find the length of side \( p \) in \( \triangle NOP \) using the Law of Cosines, we can apply the formula: \[ p^2 = n^2 + o^2 - 2no \cdot \cos(P) \] Substituting in the provided values: \[ p^2 = (3.1)^2 + (5.9)^2 - 2 \cdot (3.1) \cdot (5.9) \cdot \cos(171^\circ) \] Now calculating each term: \[ (3.1)^2 = 9.61, \quad (5.9)^2 = 34.81 \] \[ \cos(171^\circ) \approx -0.9848 \quad (\text{using a calculator}) \] \[ 2 \cdot 3.1 \cdot 5.9 \cdot (-0.9848) \approx -36.3528 \] Substituting back into the equation: \[ p^2 = 9.61 + 34.81 + 36.3528 \] \[ p^2 \approx 80.7728 \] Now taking the square root: \[ p \approx \sqrt{80.7728} \approx 8.98 \] Rounding to the nearest tenth, we find that the length of \( p \) is approximately \( 9.0 \, \text{cm} \).