In a Delta ABC,if A = 30^o,b = 2,and c = sqrt | vedclass

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(B) Using the Napier's Analogy: $	an \left( \frac{C - B}{2} ight) = \frac{c - b}{c + b} \cot \left( \frac{A}{2} ight)$.Given $A = 30^o$,$b = 2$,$

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$30$

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(B) Using the Napier's Analogy: $	an \left( \frac{C - B}{2} ight) = \frac{c - b}{c + b} \cot \left( \frac{A}{2} ight)$.Given $A = 30^o$,$b = 2$,$c = \sqrt{3} + 1$.$\frac{A}{2} = 15^o$.$\cot(15^o) = \cot(45^o - 30^o) = \frac{\cot 45^o \cot 30^o + 1}{\cot 30^o - \cot 45^o} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.Substituting the values:$	an \left( \frac{C - B}{2} ight) = \frac{(\sqrt{3} + 1) - 2}{(\sqrt{3} + 1) + 2} 	imes (2 + \sqrt{3})$$= \frac{\sqrt{3} - 1}{\sqrt{3} + 3} 	imes (2 + \sqrt{3})$$= \frac{\sqrt{3} - 1}{\sqrt{3}(\sqrt{3} + 1)} 	imes (2 + \sqrt{3})$$= \frac{(\sqrt{3} - 1)(\sqrt{3} + 1)}{\sqrt{3}(\sqrt{3} + 1)^2} 	imes (2 + \sqrt{3})$ (This simplifies to $\frac{1}{\sqrt{3}}$).Thus,$	an \left( \frac{C - B}{2} ight) = \frac{1}{\sqrt{3}} = 	an 30^o$.Therefore,$\frac{C - B}{2} = 30^o$.

In a $\Delta ABC$,if $A = 30^o$,$b = 2$,and $c = \sqrt{3} + 1$,then $\frac{C - B}{2} = $ ....$^o$

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