One side of the triangle is 10cm smaller than the other, and the angle between these sides is 60

One side of the triangle is 10cm smaller than the other, and the angle between these sides is 60 degrees. Find the larger of these sides if the third side of the triangle is 14cm.

Let the largest of the unknown sides of a given triangle be x cm, then the smaller one is (x – 10) cm.According to the cosine theorem, the square of the side of a triangle is equal to the sum of the squares of its other two sides minus their doubled product by the cosine of the angle between them. Let’s make the equation:

x ^ 2 + (x – 10) ^ 2 – 2 * x * (x – 10) * cos 60 ° = 142;

x ^ 2 + x ^ 2 + 100 – 2 * 10 * x – 2 * x * (x – 10) * 0.5 = 196;

x ^ 2 + x ^ 2 + 100 – 20x – x ^ 2 + 10x = 196;

x ^ 2 – 10x – 96 = 0.

D = 10 ^ 2 – 4 * (- 96) = 100 + 384 = 484 = 22 ^ 2;

x = (10 ± √D) / 2;

x = (10 + 22) / 2 = 16 cm – the large side of the triangle.