One side of the triangle is 8 cm larger than the other, and the angle between them is 120 °.

One side of the triangle is 8 cm larger than the other, and the angle between them is 120 °. Find the perimeter of the triangle if the third side is 28 cm.

Let’s make the equation, AC = x, then CB = (x + 8), now we can solve by the cosine theorem:
AB ^ 2 = AC ^ 2 + BC ^ 2 – 2 * AC * BC * cos 120 ° = x ^ 2 + (x + 8) ^ 2 – 2 * x * (x + 8) * (- 1/2) = x ^ 2 + x ^ 2 + 16x + 64 + x ^ 2 + 8x = 3 * x ^ 2 + 24x + 64;

Now let’s bring to the equation:
3 * x ^ 2 + 24x + 64 = 28 ^ 2;
3 * x ^ 2 + 24x + 64 = 784;
3 * x ^ 2 + 24x – 720 = 0; (/ 3);
x ^ 2 + 8x – 240 = 0;
D = 1024, √1024 = 32;
x1 = (- 8 + 32) / 2 = 12;
x2 = (- 8 – 32) / 2 = – 20;

12 cm, there will be AC.

– 20 does not satisfy the solution because it is less than 0.

BC = 12 + 8;

BC = 20 cm;

P = AB + AC + BC;

P = 28 + 12 + 20;

P = 60 cm;

Answer: P – 60 cm.