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(C) In $\Delta ABC$,the sum of angles is $180^{\circ}$. Given $\angle B = 120^{\circ}$ and $\angle C = \angle A = 	heta$,we have $120^{\circ} + 2	he

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side $a = 	ext{side } c = \frac{10 \sqrt{3}}{3} \, m$

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(C) In $\Delta ABC$,the sum of angles is $180^{\circ}$. Given $\angle B = 120^{\circ}$ and $\angle C = \angle A = 	heta$,we have $120^{\circ} + 2	heta = 180^{\circ}$,so $2	heta = 60^{\circ}$,which means $	heta = 30^{\circ}$.Draw a perpendicular $BM$ from $B$ to $AC$. Since $\Delta ABC$ is isosceles,$M$ is the midpoint of $AC$,so $CM = 5 \, m$.In right-angled $\Delta BMC$,$\cos 30^{\circ} = \frac{CM}{BC}$.$\frac{\sqrt{3}}{2} = \frac{5}{a}$.$a = \frac{10}{\sqrt{3}} = \frac{10 \sqrt{3}}{3} \, m$.Since the triangle is isosceles,$a = c = \frac{10 \sqrt{3}}{3} \, m$.

For the triangle shown in the figure,side $b = 10 \, m$,and angles $\angle C$ and $\angle A$ are equal. Find the lengths of sides $a$ and $c$.

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