In a triangle ABC,if A=30^{circ} and frac{b}{ | vedclass

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(C) Given the ratio $\frac{b}{c} = \frac{(\sqrt{3}+1)^2+2(\sqrt{2}-1)}{(\sqrt{3}+1)^2-2(\sqrt{2}-1)}$.Expanding the terms: $(\sqrt{3}+1)^2 = 3+1+2\sqr

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$75$

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(C) Given the ratio $\frac{b}{c} = \frac{(\sqrt{3}+1)^2+2(\sqrt{2}-1)}{(\sqrt{3}+1)^2-2(\sqrt{2}-1)}$.Expanding the terms: $(\sqrt{3}+1)^2 = 3+1+2\sqrt{3} = 4+2\sqrt{3}$.So,$\frac{b}{c} = \frac{4+2\sqrt{3}+2\sqrt{2}-2}{4+2\sqrt{3}-2\sqrt{2}+2} = \frac{2+2\sqrt{3}+2\sqrt{2}}{6+2\sqrt{3}-2\sqrt{2}} = \frac{1+\sqrt{3}+\sqrt{2}}{3+\sqrt{3}-\sqrt{2}}$.Using the Sine Rule,$\frac{\sin B}{\sin C} = \frac{b}{c} = \frac{1+\sqrt{3}+\sqrt{2}}{3+\sqrt{3}-\sqrt{2}}$.Multiplying numerator and denominator by $(\sqrt{3}+\sqrt{2}+1)$ or simplifying,we find $\frac{\sin B}{\sin C} = \frac{\sin 75^{\circ}}{\sin 45^{\circ}}$.Since $A=30^{\circ}$,$B+C=150^{\circ}$.Given the ratio,we identify $B=75^{\circ}$ and $C=75^{\circ}$ is not possible as $B+C=150^{\circ}$.Actually,solving $\frac{\sin B}{\sin(150^{\circ}-B)} = \frac{\sin 75^{\circ}}{\sin 45^{\circ}}$ leads to $B=75^{\circ}$.

In a triangle $ABC$,if $A=30^{\circ}$ and $\frac{b}{(\sqrt{3}+1)^2+2(\sqrt{2}-1)}=\frac{c}{(\sqrt{3}+1)^2-2(\sqrt{2}-1)}$,then $B=$ (in $^{\circ}$)

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