In a Delta ABC,frac{a}{b} = 2 + sqrt{3} and a | vedclass

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(A) Given $\frac{a}{b} = 2 + \sqrt{3}$ and $\angle C = 60^\circ$. Using the Law of Tangents: $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot\le

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$(105^\circ, 15^\circ)$

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(A) Given $\frac{a}{b} = 2 + \sqrt{3}$ and $\angle C = 60^\circ$. Using the Law of Tangents: $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}ight)$. $\frac{a-b}{a+b} = \frac{(2+\sqrt{3})b - b}{(2+\sqrt{3})b + b} = \frac{1+\sqrt{3}}{3+\sqrt{3}} = \frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}}$. $\cot\left(\frac{60^\circ}{2}ight) = \cot(30^\circ) = \sqrt{3}$. So,$	an\left(\frac{A-B}{2}ight) = \frac{1}{\sqrt{3}} 	imes \sqrt{3} = 1$. $\frac{A-B}{2} = 45^\circ \implies A-B = 90^\circ$. Since $A+B+C = 180^\circ$ and $C = 60^\circ$,$A+B = 120^\circ$. Solving $A-B = 90^\circ$ and $A+B = 120^\circ$,we get $2A = 210^\circ \implies A = 105^\circ$ and $B = 15^\circ$. Thus,the ordered pair is $(105^\circ, 15^\circ)$.

In a $\Delta ABC$,$\frac{a}{b} = 2 + \sqrt{3}$ and $\angle C = 60^\circ$. Then the ordered pair $(\angle A, \angle B)$ is equal to

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