A triangle has side lengths that are integers, one of which is 7, and the measure of the angle opposite to it is $60^{\circ}$. What could be the area of the triangle?
Solution. The conditions of the problem are clearly satisfied by the triangle in which the other two sides are also 7. In this case, the area of the equilateral triangle is

$$
T=\frac{49 \sqrt{3}}{4} \approx 21.22
$$

Construct the $60^{\circ}$ visual circle over the 7-unit $AB$ segment. Take a point $C$ on the circle such that $\angle ACB = 60^{\circ}$. The diameter of the circle is:

$$
\frac{\frac{7}{2}}{r}=\sin 60^{\circ}=\frac{\sqrt{3}}{2} \Rightarrow 2r=\frac{7 \cdot 2}{\sqrt{3}} \approx 8.08
$$

Therefore, the lengths of sides $AC$ and $BC$ (since they are integers) are at most 8. Since we can assume that the triangle is not equilateral, i.e., it does not have two $60^{\circ}$ angles, it cannot have two sides of length 7 either. The sum of the angles opposite to sides $BC$ and $AC$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$, so the larger one is greater than $60^{\circ}$, and thus the opposite side is longer than 7; since the length is an integer and at most 8, the longer side, e.g., $BC = 8$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9c1a44d41c41bec8e583g-1.jpg?height=493&width=501&top_left_y=716&top_left_x=801)

Let's apply the cosine theorem to triangle $ABC$:

$$
49=64+b^{2}-2 \cdot 8 \cdot b \cdot \cos 60^{\circ}
$$

Since $\cos 60^{\circ}=\frac{1}{2}$, we get the following quadratic equation for $b$:

$$
b^{2}-8 b+15=0, \quad \text { hence } \quad b_{1}=5 \text { and } b_{2}=3
$$

Thus, two triangles satisfy the conditions, with sides 8, 7, and 5, and the area of the first triangle is:

$$
T_{1}=\frac{8 \cdot 5 \cdot \sin 60^{\circ}}{2}=\frac{20 \sqrt{3}}{2} \approx 17.32
$$

The sides of the second triangle are 8, 7, and 3, and its area is:

$$
T_{2}=\frac{8 \cdot 3 \cdot \sqrt{3}}{4}=6 \cdot \sqrt{3} \approx 10.39
$$

The problem has a total of three solutions.