Find the perimeter of the triangle ABC if AB = 19, AC = 5, angle C = 120 °.

1. Consider the triangle ABC, <C = 120 °, AB = 19, AC = 5, find its perimeter:
The perimeter of the triangle is:
p = a + b + c = AB + AC + CB.
2. To determine the side СB, we use the cosine theorem:
AB² = CB² + AC²-2 * CB * AC * cos C.
Substitute the numbers we know, and denote CB by x:
19² = x² + 5²-2 * x * 5 * cos C.
19² = x² + 5²-2 * x * 5 * (- 1/2).
(-1/2) = – 19² + x² + 5² / (2 * x * 5).
-10x = 2x²-672.
2x² + 10x-672 = 0
D = b-4ac = (- 5) ²-4 * 1 (-336) = 1369.
X1 = -5-√1369 / (2 * 1) = – 21.
x2 = -5 + √1369 / (2 * 1) = 16.
The length cannot be negative, so only CB = 16 cm satisfies us.
3.Let’s define the perimeter:
p = AB + AC + CB = 19 + 5 + 16 = 40 cm.
Answer: the perimeter is 40 cm.