The sides of a triangle are 9.10.11 cm. Find the cosine of the larger angle of this triangle.

In order to calculate the cosine of the larger angle of a given triangle, it is necessary to apply the cosine theorem, according to which the square of one side of a triangle is equal to the sum of the squares of its other two sides minus the doubled product of these sides by the cosine of the angle between them:

a ^ 2 = b ^ 2 + c ^ 2 – 2bc * cos α.

The major angle of a given triangle is the angle opposite to the larger side. Let’s assume that:

AB = 9 cm, BC = 10 cm, AC = 11 cm. Therefore, the large angle of this triangle is the angle ∠B.

AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 AB BC cos B;

cos B = (AB ^ 2 + BC ^ 2 – AC ^ 2) / 2 AB BC;

cos B = (9 ^ 2 + 10 ^ 2 – 11 ^ 2) / 2 * 9 * 10 = (81 + 100 – 121) / 180 = 60/180 = 1/3.

Answer: The cosine of the larger angle is 1/3.