If in a triangle ABC,b = sqrt{3},c = 1 and B | vedclass

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(A) Using the Napier's analogy: $	an \left( \frac{B - C}{2} ight) = \frac{b - c}{b + c} \cot \left( \frac{A}{2} ight)$.Given $B - C = 90^{\circ}$

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$30$

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(A) Using the Napier's analogy: $	an \left( \frac{B - C}{2} ight) = \frac{b - c}{b + c} \cot \left( \frac{A}{2} ight)$.Given $B - C = 90^{\circ}$,$b = \sqrt{3}$,and $c = 1$.Substituting the values: $	an \left( \frac{90^{\circ}}{2} ight) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cot \left( \frac{A}{2} ight)$.Since $	an(45^{\circ}) = 1$,we have $1 = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cot \left( \frac{A}{2} ight)$.$\cot \left( \frac{A}{2} ight) = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$.Rationalizing the denominator: $\frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.Since $\cot(15^{\circ}) = 2 + \sqrt{3}$,we have $\frac{A}{2} = 15^{\circ}$.Therefore,$A = 30^{\circ}$.

If in a triangle $ABC$,$b = \sqrt{3}$,$c = 1$ and $B - C = 90^{\circ}$,then $\angle A$ is .....$^{\circ}$.

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