z_1, z_2, z_3 represent the vertices A, B, C | vedclass

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(B) Let $a, b, c$ be the side lengths $BC, AC, AB$ respectively.Given $|z_1-z_2| = c = \sqrt{25-12\sqrt{3}}$.Given $\frac{|z_1-z_3|}{|z_2-z_3|} = \fra

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$3$

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(B) Let $a, b, c$ be the side lengths $BC, AC, AB$ respectively.Given $|z_1-z_2| = c = \sqrt{25-12\sqrt{3}}$.Given $\frac{|z_1-z_3|}{|z_2-z_3|} = \frac{b}{a} = \frac{3}{4}$,so $b = \frac{3}{4}a$.In $	riangle ABC$,by the Law of Cosines:$c^2 = a^2 + b^2 - 2ab \cos(30^{\circ})$$25-12\sqrt{3} = a^2 + (\frac{3}{4}a)^2 - 2a(\frac{3}{4}a)(\frac{\sqrt{3}}{2})$$25-12\sqrt{3} = a^2 + \frac{9}{16}a^2 - \frac{3\sqrt{3}}{4}a^2$$25-12\sqrt{3} = a^2(\frac{16+9-12\sqrt{3}}{16}) = a^2(\frac{25-12\sqrt{3}}{16})$Thus,$a^2 = 16$,so $a = 4$.Then $b = \frac{3}{4}(4) = 3$.The area of $	riangle ABC = \frac{1}{2}ab \sin(30^{\circ}) = \frac{1}{2}(4)(3)(\frac{1}{2}) = 3$ sq. units.

$z_1, z_2, z_3$ represent the vertices $A, B, C$ of a triangle $ABC$ respectively in the Argand plane. If $|z_1-z_2|=\sqrt{25-12\sqrt{3}}$,$|\frac{z_1-z_3}{z_2-z_3}|=\frac{3}{4}$ and $\angle ACB=30^{\circ}$,then the area (in sq. units) of that triangle is

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