In triangle ABC,if a=1, b=2, angle C=60^{circ | vedclass

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(A) Given: $a=1, b=2, \angle C=60^{\circ}$.$1$. Calculate the area $\Delta$ of the triangle:$\Delta = \frac{1}{2} ab \sin C$$\Delta = \frac{1}{2} 	im

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$6$

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(A) Given: $a=1, b=2, \angle C=60^{\circ}$.$1$. Calculate the area $\Delta$ of the triangle:$\Delta = \frac{1}{2} ab \sin C$$\Delta = \frac{1}{2} 	imes 1 	imes 2 	imes \sin 60^{\circ}$$\Delta = 1 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$\Delta^2 = \frac{3}{4}$$4 \Delta^2 = 3$$2$. Calculate the side $c$ using the Law of Cosines:$c^2 = a^2 + b^2 - 2ab \cos C$$c^2 = 1^2 + 2^2 - 2(1)(2) \cos 60^{\circ}$$c^2 = 1 + 4 - 4 	imes \frac{1}{2}$$c^2 = 5 - 2 = 3$$3$. Calculate $4 \Delta^2 + c^2$:$4 \Delta^2 + c^2 = 3 + 3 = 6$.

In $\triangle ABC$,if $a=1, b=2, \angle C=60^{\circ}$,then find the value of $4 \Delta^2+c^2$.

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