Question: Determine the measure of angle $B$ in the following diagram: Triangle $ABC$, with vertices $A$ at the top, $B$ at the bottom-left, and $C$ at the right. Side $c = 9.7$ cm is the left side from $B$ to $A$, side $b = 9.0$ cm is the right side from $A$ to $C$, and side $a = 12.7$ cm is the bottom side from $B$ to $C$.

1. **Problem Statement:** We need to find the measure of angle $B$ in triangle $ABC$ where: - Side $a = 12.7$ cm (opposite angle $A$) - Side $b = 9.0$ cm (opposite angle $B$) - Side $c = 9.7$ cm (opposite angle $C$) 2. **Formula Used:** We use the Law of Cosines to find angle $B$: $$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$$ This formula relates the lengths of the sides of a triangle to the cosine of one of its angles. 3. **Substitute the known values:** $$\cos B = \frac{12.7^2 + 9.7^2 - 9.0^2}{2 \times 12.7 \times 9.7}$$ Calculate each term: $$12.7^2 = 161.29$$ $$9.7^2 = 94.09$$ $$9.0^2 = 81.00$$ So, $$\cos B = \frac{161.29 + 94.09 - 81.00}{2 \times 12.7 \times 9.7} = \frac{174.38}{246.38}$$ 4. **Simplify the fraction:** $$\cos B = \frac{\cancel{174.38}}{\cancel{246.38}} \approx 0.7075$$ 5. **Find angle $B$ by taking the inverse cosine:** $$B = \cos^{-1}(0.7075)$$ Using a calculator, $$B \approx 45.0^\circ$$ **Final answer:** The measure of angle $B$ is approximately **$45.0^\circ$**.