In the triangle ABC, the angle C is straight, the leg BC = 12, and the cosine of the acute angle ABC

In the triangle ABC, the angle C is straight, the leg BC = 12, and the cosine of the acute angle ABC is 0.8. Find the area of the triangle.

If the angle C is straight in the triangle ABC, then we are dealing with a right-angled triangle, where AB is the hypotenuse, AC and BC are the legs.
According to the definition of cosine, cos∠ABC = BC / AB. Substituting these values, we have: 0.8 = 12 / AB, whence AB = 12: 0.8 = 15.
By the Pythagorean theorem AB^2 = AC^2 + BC^2. Therefore, AC ^ 2 = AB ^ 2 – BC ^ 2 = 15 ^ 2 – 12 ^ 2 = 225 – 144 = 81, whence AC = 9.
Since BC is the height of the right-angled triangle ABC, its area (S) is found by the formula S = ½ * AC * BC. We have S = ½ * 9 * 12 = 54.
Answer: 54.