1. **State the problem:** Given a triangle with angle $\angle A = 57^\circ$, side $c = 27$ m, and side $a = 25$ m, find the missing parts of the triangle. 2. **Formula and rules:** Use the Law of Cosines to find side $b$: $$b^2 = a^2 + c^2 - 2ac \cos(\angle A)$$ Then use the Law of Sines to find the other angles: $$\frac{\sin(\angle B)}{b} = \frac{\sin(\angle A)}{a}$$ 3. **Calculate side $b$ using Law of Cosines:** $$b^2 = 25^2 + 27^2 - 2 \times 25 \times 27 \times \cos(57^\circ)$$ $$b^2 = 625 + 729 - 1350 \times \cos(57^\circ)$$ Calculate $\cos(57^\circ) \approx 0.5446$: $$b^2 = 1354 - 1350 \times 0.5446 = 1354 - 735.21 = 618.79$$ $$b = \sqrt{618.79} \approx 24.88 \text{ m}$$ 4. **Find angle $B$ using Law of Sines:** $$\frac{\sin(\angle B)}{b} = \frac{\sin(57^\circ)}{25}$$ $$\sin(\angle B) = \frac{b \times \sin(57^\circ)}{25} = \frac{24.88 \times 0.8387}{25} = 0.8347$$ $$\angle B = \arcsin(0.8347) \approx 56.3^\circ$$ 5. **Find angle $C$:** $$\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 57^\circ - 56.3^\circ = 66.7^\circ$$ **Final answers:** - Side $b \approx 24.88$ m - Angle $B \approx 56.3^\circ$ - Angle $C \approx 66.7^\circ$