# The resultant of two forces applied to the body at an angle of 120 degrees is equal to 100 N.

**The resultant of two forces applied to the body at an angle of 120 degrees is equal to 100 N. What is one of the components of these forces if the other component is equal to 100 N?**

Let given: the resultant F = 100 N of two forces F₁ = 100 N and F₂, applied to the body at an angle α = 120 °. To determine what the second component F₂ of these forces is equal to, we use the parallelogram rule. In the parallelogram ABCD, the body is at the vertex A, side AB = F₁, AD = F₂, ∠BAD = 120 °, then the diagonal AC = F. In isosceles ΔABC (since AB = AC), the angles at the base of BC are equal to x, ∠BAC = 120 ° – x, since ∠САD = ∠АСВ = x (as lying at BC | | AD and secant AC). Knowing that the sum of the angles in a triangle is 180 °, we get the equation:

x + x + 120 ° – x = 180 °;

x = 30 °; then,

∠ВАС = 120 ° – 30 ° = 90 °;

by the Pythagorean theorem BC² = AC² + AB²;

ВС² = 100² + 100²;

BC = 100 √2, but BC = AD, then

F₂ = 100 √2 ≈ 141 (H).

Answer: the second component of forces ≈ 141 N.