If two sides of a triangle are sqrt{3}-2 and | vedclass

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(D) Let the sides of the triangle be $a = \sqrt{3}-2$ and $b = \sqrt{3}+2$,and the included angle be $C = 60^{\circ}$.Using the Law of Cosines,the thi

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$\sqrt{15}$ units

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(D) Let the sides of the triangle be $a = \sqrt{3}-2$ and $b = \sqrt{3}+2$,and the included angle be $C = 60^{\circ}$.Using the Law of Cosines,the third side $c$ is given by:$c^2 = a^2 + b^2 - 2ab \cos(C)$$c^2 = (\sqrt{3}-2)^2 + (\sqrt{3}+2)^2 - 2(\sqrt{3}-2)(\sqrt{3}+2) \cos(60^{\circ})$$c^2 = (3 + 4 - 4\sqrt{3}) + (3 + 4 + 4\sqrt{3}) - 2(3 - 4) 	imes \frac{1}{2}$$c^2 = 7 - 4\sqrt{3} + 7 + 4\sqrt{3} - 2(-1) 	imes \frac{1}{2}$$c^2 = 14 + 1 = 15$$c = \sqrt{15}$ units.

If two sides of a triangle are $\sqrt{3}-2$ and $\sqrt{3}+2$ units and their included angle is $60^{\circ}$,then the third side of the triangle is

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