In a Delta ABC,if b = a(sqrt{3} - 1) and angl | vedclass

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(D) Given $b = a(\sqrt{3} - 1)$ and $\angle C = 30^o$. Since $\angle A + \angle B + \angle C = 180^o$,we have $\angle A + \angle B = 150^o$. Using the

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$105$

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(D) Given $b = a(\sqrt{3} - 1)$ and $\angle C = 30^o$. Since $\angle A + \angle B + \angle C = 180^o$,we have $\angle A + \angle B = 150^o$. Using the Napier's Analogy: $	an\left(\frac{A - B}{2}ight) = \frac{a - b}{a + b} \cot\left(\frac{C}{2}ight)$. Substituting $b = a(\sqrt{3} - 1)$: $\frac{a - a(\sqrt{3} - 1)}{a + a(\sqrt{3} - 1)} = \frac{1 - \sqrt{3} + 1}{1 + \sqrt{3} - 1} = \frac{2 - \sqrt{3}}{\sqrt{3}}$. Also,$\cot\left(\frac{30^o}{2}ight) = \cot(15^o) = 2 + \sqrt{3}$. So,$	an\left(\frac{A - B}{2}ight) = \left(\frac{2 - \sqrt{3}}{\sqrt{3}}ight)(2 + \sqrt{3}) = \frac{4 + 2\sqrt{3} - 2\sqrt{3} - 3}{\sqrt{3}} = \frac{1}{\sqrt{3}}$. Thus,$\frac{A - B}{2} = 30^o$,which implies $A - B = 60^o$. Solving $A + B = 150^o$ and $A - B = 60^o$,we get $2A = 210^o$,so $A = 105^o$.

In a $\Delta ABC$,if $b = a(\sqrt{3} - 1)$ and $\angle C = 30^o$,then the measure of the angle $A$ is.....$^o$.

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