## Task B-4.2.

The lengths of the sides of a triangle form an arithmetic sequence with a difference of 2. If the sine of the smallest angle in this triangle is equal to $\frac{\sqrt{15}}{8}$, calculate the perimeter of this triangle.
## Solution.

Let the sides of the triangle be of length $n, n+2$, and $n+4$.

Denote by $\alpha$ the smallest angle of the triangle, opposite the shortest side of length $n$.

According to the cosine rule, we have

$n^{2}=(n+2)^{2}+(n+4)^{2}-2(n+2)(n+4) \cos \alpha$.

Since $\sin \alpha=\frac{\sqrt{15}}{8}$, it follows that $\cos \alpha= \pm \sqrt{1-\frac{15}{64}}= \pm \sqrt{\frac{49}{64}}= \pm \frac{7}{8}$.

Since $\alpha$ is the smallest angle of the triangle, it is certain that $\alpha<90^{\circ}$, so $\cos \alpha=\frac{7}{8}$.

Solve the equation:

$$
\begin{aligned}
& n^{2}=(n+2)^{2}+(n+4)^{2}-2(n+2)(n+4) \cdot \frac{7}{8} \\
& n^{2}=n^{2}+4 n+4+n^{2}+8 n+16-\left(n^{2}+6 n+8\right) \cdot \frac{7}{4} \\
& 0=4 n^{2}+48 n+80-7 n^{2}-42 n-56 \\
& 3 n^{2}-6 n-24=0 \\
& n^{2}-2 n-8=0 \\
& n_{1}=4, \quad n_{2}=-2
\end{aligned}
$$

The lengths of the sides of the triangle are 4, 6, and 8.

The perimeter of the triangle is 18.

Note: If the lengths of the sides of the triangle are denoted by $(n-2), n, (n+2)$, then the quadratic equation derived from the cosine rule is $3 n^{2}-18 n=0$ and its solutions are $n_{1}=0, n_{2}=6$.