## Task B-3.7.

A children's playground has the shape of a triangle with a perimeter of $36 \mathrm{~m}$ and an area of $30 \sqrt{3} \mathrm{~m}^{2}$. If the measure of one of the angles is $60^{\circ}$, determine the lengths of the sides of the playground.
## First solution.

Let the lengths of the sides of the triangle be $a, b$, and $c$, and the given angle of $60^{\circ}$ is between the sides of lengths $a$ and $b$.

From the formula for the area of a triangle, we have $P=\frac{1}{2} a b \sin 60^{\circ}=30 \sqrt{3}$, which implies $a b=120$.

From the given perimeter, we get $o=a+b+c=36$.

We use the cosine rule for side $c$.

$$
c^{2}=a^{2}+b^{2}-2 a b \cos 60^{\circ}=a^{2}+b^{2}-a b
$$

By squaring the equation $c=36-(a+b)$, we get

$$
c^{2}=36^{2}-72(a+b)+(a+b)^{2}
$$

so

$$
a^{2}+b^{2}-a b=36^{2}-72(a+b)+\left(a^{2}+2 a b+b^{2}\right)
$$

which simplifies to $36^{2}-72(a+b)+3 a b=0$.

Since $a b=120$, we have

$$
36^{2}-72(a+b)+360=0
$$

which implies $a+b=23$. Therefore, $c=36-23=13$.

From $a b=120$ and $a+b=23$, it follows that, by Vieta's formulas, $a$ and $b$ are the solutions of the quadratic equation $x^{2}-23 x+120=0$.

These are $a=15$ and $b=8$ (or vice versa).

The sides of the playground are $8 \mathrm{~m}, 13 \mathrm{~m}$, and $15 \mathrm{~m}$ long.