Let ABC and ABC^{prime} be two non-congruent | vedclass

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(C) Using the cosine rule in $	riangle ABC$ and $	riangle ABC^{\prime}$ with $AB=c=4$,$AC=AC^{\prime}=b=2\sqrt{2}$,and $\angle B=30^{\circ}$:$\cos 3

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$4$

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(C) Using the cosine rule in $	riangle ABC$ and $	riangle ABC^{\prime}$ with $AB=c=4$,$AC=AC^{\prime}=b=2\sqrt{2}$,and $\angle B=30^{\circ}$:$\cos 30^{\circ} = \frac{a^2+c^2-b^2}{2ac} \Rightarrow \frac{\sqrt{3}}{2} = \frac{a^2+16-8}{2 \cdot a \cdot 4}$$\Rightarrow 4\sqrt{3}a = a^2+8$ $\Rightarrow a^2 - 4\sqrt{3}a + 8 = 0$Let $a_1$ and $a_2$ be the two possible lengths of side $BC$. From the quadratic equation,$a_1+a_2 = 4\sqrt{3}$ and $a_1a_2 = 8$.The difference $|a_1-a_2| = \sqrt{(a_1+a_2)^2 - 4a_1a_2} = \sqrt{(4\sqrt{3})^2 - 4(8)} = \sqrt{48-32} = \sqrt{16} = 4$.The area of a triangle is given by $\Delta = \frac{1}{2}ac \sin B$.The difference in areas is $|\Delta_1 - \Delta_2| = |\frac{1}{2}a_1c \sin B - \frac{1}{2}a_2c \sin B| = \frac{1}{2}c \sin B |a_1-a_2|$.Substituting the values: $|\Delta_1 - \Delta_2| = \frac{1}{2} \cdot 4 \cdot \sin 30^{\circ} \cdot 4 = \frac{1}{2} \cdot 4 \cdot \frac{1}{2} \cdot 4 = 4$.

Let $ABC$ and $ABC^{\prime}$ be two non-congruent triangles with sides $AB=4$,$AC=AC^{\prime}=2\sqrt{2}$ and $\angle B=30^{\circ}$. The absolute value of the difference between the areas of these triangles is

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