The area (in square units) of triangle ABC if | vedclass

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(D) Given,$\angle A=75^{\circ}, \angle B=45^{\circ}$ and $a=2(\sqrt{3}+1)$.In $	riangle AOC$,$	an 60^{\circ} = \frac{x}{y}$ $\Rightarrow \sqrt{3} =

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$6+2\sqrt{3}$

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(D) Given,$\angle A=75^{\circ}, \angle B=45^{\circ}$ and $a=2(\sqrt{3}+1)$.In $	riangle AOC$,$	an 60^{\circ} = \frac{x}{y}$ $\Rightarrow \sqrt{3} = \frac{x}{y}$ $\Rightarrow x = \sqrt{3}y$.Now,$x+y = 2(\sqrt{3}+1)$.Substituting $x = \sqrt{3}y$,we get $\sqrt{3}y + y = 2(\sqrt{3}+1)$ $\Rightarrow y(\sqrt{3}+1) = 2(\sqrt{3}+1)$ $\Rightarrow y = 2$.Then,$x = 2\sqrt{3}$.Now,the area of $	riangle ABC = 	ext{Area of } 	riangle AOB + 	ext{Area of } 	riangle AOC$.Area $= \frac{1}{2} 	imes x 	imes x + \frac{1}{2} 	imes x 	imes y = \frac{1}{2}x(x+y)$.Area $= \frac{1}{2} 	imes (2\sqrt{3}) 	imes (2\sqrt{3} + 2) = \sqrt{3} 	imes 2(\sqrt{3}+1) = 2(3 + \sqrt{3}) = 6 + 2\sqrt{3} 	ext{ sq units}$.

The area (in square units) of $\triangle ABC$ if $\angle A=75^{\circ}, \angle B=45^{\circ}$ and $a=2(\sqrt{3}+1)$ is

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