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Question Number 41027 by behi83417@gmail.com last updated on 31/Jul/18

	Answered by MJS last updated on 01/Aug/18

	$A = B \land a = b$ $2 a \tan A = c \tan (π - 2 A)$ $2 a \tan A = - c \tan 2 A$ $let c = 1$ $- 2 a = \frac{\tan 2 A}{\tan A}$ $a = \frac{\cos^{2} A}{1 - {2 \cos}^{2} A}$ $\cos^{2} A = \frac{a}{2 a + 1}$ $\frac{\sin A}{1} = \frac{\sin (π - 2 A)}{a}$ $\sin A = \frac{\sin 2 A}{a}$ $a = \frac{\sin 2 A}{\sin A}$ $a = 2 \cos A$ $\cos A = \frac{a}{2}$ $\frac{a}{2 a + 1} = \frac{a^{2}}{4}$ $a_{1} = 0$ $a^{2} + \frac{1}{2} a - 2 = 0$ $a_{2} = - \frac{1}{4} - \frac{\sqrt{33}}{4}$ $a_{3} = a = - \frac{1}{4} + \frac{\sqrt{33}}{4} \approx . 5931$ $\cos A = - \frac{1}{8} + \frac{\sqrt{33}}{8} \Rightarrow A = \arccos (- \frac{1}{8} + \frac{\sqrt{33}}{8}) \approx 53.62 °$ $\Rightarrow C = 2 \arcsin (- \frac{1}{8} + \frac{\sqrt{33}}{8}) \approx 72.75 °$
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