1. **Problem statement:** Solve triangle ABC given $A=43^\circ$, $b=7$ cm, and $c=6$ cm. 2. **Known values:** - Angle $A=43^\circ$ - Side $b=7$ cm (opposite angle $B$) - Side $c=6$ cm (opposite angle $C$) 3. **Goal:** Find angles $B$, $C$ and side $a$. 4. **Formula used:** Law of Cosines and Law of Sines. 5. **Step 1: Find side $a$ using Law of Cosines:** $$a^2 = b^2 + c^2 - 2bc \cos A$$ Substitute values: $$a^2 = 7^2 + 6^2 - 2 \times 7 \times 6 \times \cos 43^\circ$$ Calculate: $$a^2 = 49 + 36 - 84 \times 0.7314 = 85 - 61.4376 = 23.5624$$ $$a = \sqrt{23.5624} \approx 4.854 \text{ cm}$$ 6. **Step 2: Find angle $B$ using Law of Sines:** $$\frac{a}{\sin A} = \frac{b}{\sin B}$$ Rearranged: $$\sin B = \frac{b \sin A}{a} = \frac{7 \times \sin 43^\circ}{4.854}$$ Calculate: $$\sin B = \frac{7 \times 0.6820}{4.854} = \frac{4.774}{4.854} = 0.9839$$ $$B = \arcsin(0.9839) \approx 78.5^\circ$$ 7. **Step 3: Find angle $C$ using angle sum:** $$C = 180^\circ - A - B = 180^\circ - 43^\circ - 78.5^\circ = 58.5^\circ$$ 8. **Final answers:** - $a \approx 4.85$ cm - $B \approx 78.5^\circ$ - $C \approx 58.5^\circ$