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.

We are given triangle NOP where side n = 3.1 cm, side o = 5.9 cm, and angle P = 171°. Recall that in triangle NOP, side p is opposite angle P. Using the Law of Cosines:   p² = n² + o² – 2·n·o·cos(P) Step 1: Substitute the given values:   p² = (3.1)² + (5.9)² – 2·(3.1)·(5.9)·cos(171°) Step 2: Compute the squares:   (3.1)² = 9.61   (5.9)² = 34.81   9.61 + 34.81 = 44.42 Step 3: Evaluate the cosine: Remember that cos(171°) = cos(180° − 9°) = –cos(9°). With cos(9°) approximately 0.9877, we have:   cos(171°) ≈ –0.9877 Step 4: Substitute and compute:   p² = 44.42 – 2·(3.1)·(5.9)·(–0.9877) Since the double negative becomes positive:   2·(3.1)·(5.9) = 36.58   36.58 × 0.9877 ≈ 36.14 Thus:   p² ≈ 44.42 + 36.14 = 80.56 Step 5: Take the square root to find p:   p ≈ √80.56 ≈ 8.98 cm Rounded to the nearest tenth, the length of p is approximately 9.0 cm. Final Answer: p ≈ 9.0 cm.