One of the sides of the triangle is 11 cm larger than the other, the angle between them is 120 degrees

One of the sides of the triangle is 11 cm larger than the other, the angle between them is 120 degrees and the third side is 19 cm. Find the perimeter and area of the triangle

Let one of the sides of the triangle be x cm, then the second is (x + 11) cm. The square of the third side is equal to the sum of the squares of the first and second sides minus their doubled product by the cosine of the angle between them. Let’s make the equation:

x ^ 2 + (x + 11) ^ 2 – 2 * x * (x + 11) * cos 120 ° = 19 ^ 2;

x ^ 2 + x ^ 2 + 121 + 2 ^ 2 * x + x ^ 2 + 11 * x = 361;

3 * x ^ 2 + 33 * x – 240 = 0;

x ^ 2 + 11 * x – 80 = 0.

D = 112 – 4 * (- 80) = 121 + 320 = 441 = 212.

x = (- 11 + 21) / 2 = 10/2 = 5 cm – the first side of the triangle.

x + 11 = 5 + 11 = 16 cm – the second side of the triangle.

The perimeter of a triangle is equal to the sum of the lengths of its sides:

P = 19 + 5 + 16 = 40 cm.

The area of ​​a triangle is equal to half of the product of side and sine of the angle between them:

S = 0.5 * 5 * 16 * sin 120 ° = 0.5 * 5 * 16 * √3 / 2 = 20√3 ≈ 34.64 cm2.