<s>[INST] If one angle of a triangle be $60^\circ,$ the area $10\sqrt3$ sq cm.,and the perimeter $20$ cm,find the lengths of the sides. If one angle of a triangle be $60^\circ$, the area $10\sqrt3\ \mbox{cm}^2$, and the perimeter $20\ \mbox{cm}$, find the lengths of the sides.

Let $\angle A=60^\circ$ and $\frac{1}{2}bc\sin A=10\sqrt3$
$$bc=40,\quad a+b+c=20$$
$$2R\sin A+2R\sin B+2R\sin C=20$$
I am stuck here.
 [/INST] You have $(b+c)^2=(20-a)^2$ and so $b^2+c^2-bc=a^2-40a+400-120$. But by the cosine formula that is $a^2$, so you have $40a=280$ and hence $a=7$, so $b+c=13,bc=40$. Hence $b,c=5,8$.
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