The sides of the triangle are 8, 10, and 12 cm. Find the corner opposite the smaller side.

Take advantage of the fact that in any triangle the smallest of the angles lies opposite the smallest of the sides of the given triangle.

In the wording of the condition for this task, it is reported that the lengths of the sides of this triangle are 8, 10, and 12 cm.

Therefore, the length of the smallest side is 8 cm.

Let us denote the value of the angle lying opposite the side 8 cm long through x.

Using the cosine theorem, we can write the following equation:

10 ^ 2 + 12 ^ 2 – 2 * 10 * 12 * cosx = 8 ^ 2,

solving which, we get:

100 + 144 – 240 * cosx = 64;

244 – 240 * cosx = 64;

240 * cosx = 244 – 64;

240 * cosx = 180;

cosx = 180/240 = 3/4;

x = arccos (3/4).

Answer: arccos (3/4).