If one angle of a triangle be $60^\circ$, the area $10 \sqrt{3} \; cm^2$ and the perimeter $20 \; cm$, find the lengths of the sides.
Let the sides of the triangle be $a$, $b$ and $c$.
Let $\;$ $A = 60^\circ$
Given: $\;$ Area of $\triangle ABC = \Delta = 10 \sqrt{3} \; cm^2$
Now, $\;$ $\Delta = \dfrac{1}{2}b c \sin A$
i.e. $\;$ $10 \sqrt{3} = \dfrac{1}{2}b c \sin \left(60^\circ\right)$
$\implies$ $bc = \dfrac{20 \sqrt{3}}{\sqrt{3} / 2}$ $\implies$ $bc = 40$ $\;\;\; \cdots \; (1)$
Given: $\;$ Perimeter of $\triangle ABC = 20 \; cm$
Perimeter of $\triangle ABC$ $= a + b + c$
$\therefore \;$ $a + b + c = 20$ $\implies$ $b + c = 20 - a$ $\;\;\; \cdots \; (2)$
By cosine rule, we have,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$
$\begin{aligned}
i.e. \; a^2 & = b^2 + c^2 - 2 b c \cos A \\\\
& = b^2 + c^2 - 2 b c \cos \left(60^\circ\right) \;\;\; \left[\because \; A = 60^\circ\right] \\\\
& = b^2 + c^2 - 2 \times b c \times \dfrac{1}{2} \\\\
& = b^2 + c^2 - bc \\\\
& = \left(b^2 + c^2 + 2 b c\right) - 3 b c \\\\
i.e. \; a^2 & = \left(b + c\right)^2 - 3 b c \;\;\; \cdots \; (3)
\end{aligned}$
In view of equations $(1)$ and $(2)$, equation $(3)$ becomes,
$a^2 = \left(20 - a\right)^2 - 3 \times 40$
i.e. $\;$ $a^2 = 400 + a^2 - 40 a - 120$
i.e. $\;$ $40 a = 280$ $\implies$ $a = 7$
Substituting the value of $a$ in equation $(2)$ we get,
$b + c = 20 - 7 = 13$ $\implies$ $b = 13 - c$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(1)$ becomes,
$\left(13 - c\right)c = 40$
i.e. $\;$ $c^2 - 13 c + 40 = 0$
i.e. $\;$ $\left(c - 8\right) \left(c - 5\right) = 0$
$\implies$ $c = 8$ $\;$ or $\;$ $c = 5$
Substituting the value of $c$ in equation $(1)$ we get,
When $c = 8$, $\;$ $b = \dfrac{40}{c} = 5$
When $c = 5$, $\;$ $b = \dfrac{40}{c} = 8$
$\therefore \;$ The lengths of the sides of the triangle are:
$a = 7 \; cm$, $b = 5 \; cm$, $c = 8 \; cm$ $\;$ or $\;$ $a = 7 \; cm$, $b = 8 \; cm$, $c = 5 \; cm$