1. Problem 1: Given a triangle with sides $a=\sqrt{3}$, $b=\sqrt{2}$, and angle $a=60^\circ$, find angle $B$. 2. Use the Law of Cosines formula: $$b^2 = a^2 + c^2 - 2ac \cos(B)$$ but since we want angle $B$ opposite side $b$, we use the Law of Cosines rearranged for $B$: $$\cos(B) = \frac{a^2 + c^2 - b^2}{2ac}$$ 3. However, we are missing side $c$. Since the problem states angles $a, B, y$ and sides $a, b, c$ correspondingly, and angle $a=60^\circ$, side $a=\sqrt{3}$, side $b=\sqrt{2}$, we can use the Law of Sines to find $B$: $$\frac{a}{\sin(a)} = \frac{b}{\sin(B)}$$ 4. Substitute known values: $$\frac{\sqrt{3}}{\sin(60^\circ)} = \frac{\sqrt{2}}{\sin(B)}$$ 5. Calculate $\sin(60^\circ) = \frac{\sqrt{3}}{2}$: $$\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = \frac{\sqrt{2}}{\sin(B)}$$ 6. Simplify left side: $$2 = \frac{\sqrt{2}}{\sin(B)}$$ 7. Solve for $\sin(B)$: $$\sin(B) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$ 8. Find angle $B$: $$B = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ$$ --- 9. Problem 2: Given $a=4$, angle $y=60^\circ$, and side $b$ missing, find side $c$. 10. Since $b$ is missing, assume the problem meant $b$ is known or use Law of Cosines or Law of Sines accordingly. Assuming $b$ is missing and we want $c$, we need more information. If $b$ is missing, we cannot solve for $c$ directly. 11. If $b$ is given or angle $B$ is given, use Law of Cosines: $$c^2 = a^2 + b^2 - 2ab \cos(y)$$ 12. Without $b$, we cannot proceed. Assuming $b$ is given or $b=4$ (typo), then: $$c^2 = 4^2 + b^2 - 2 \times 4 \times b \times \cos(60^\circ)$$ 13. Since $\cos(60^\circ) = 0.5$, $$c^2 = 16 + b^2 - 4b$$ 14. Without $b$, $c$ cannot be found. Please provide $b$ or angle $B$. Final answers: 1) $B = 45^\circ$ 2) Cannot solve for $c$ without additional information.