8-2 Classw 1. In \( \triangle Q R S, q=93 \mathrm{~cm}, r=77 \mathrm{~cm} \) and \( \angle S=56^{\circ} \). Find the length of \( s \), to the nearest centimeter.

We are given triangle QRS with sides:   q = 93 cm (opposite vertex Q),   r = 77 cm (opposite vertex R),   and angle S = 56° (at vertex S). Since side s is opposite angle S, we can use the Law of Cosines:   s² = q² + r² − 2qr cos(S) Step 1. Substitute the given values:   s² = 93² + 77² − 2 · 93 · 77 · cos(56°) Step 2. Calculate the individual parts:   93² = 8649   77² = 5929   So, 93² + 77² = 8649 + 5929 = 14578 Also, calculate the product:   2 · 93 · 77 = 14322 Step 3. Express the formula:   s² = 14578 − 14322 · cos(56°) Using a calculator, cos(56°) ≈ 0.5592. Therefore,   s² ≈ 14578 − 14322 × 0.5592 ≈ 14578 − 8009 ≈ 6569 Step 4. Take the square root to find s:   s ≈ √6569 ≈ 81 cm Thus, the length of s is approximately 81 centimeters when rounded to the nearest centimeter.