A) Vector A
The x-component of a vector can be found by using the formula

where
v is the magnitude of the vector
is the angle between the x-axis and the direction of the vector
- Vector A has a magnitude of 50 units along the positive x-direction, so
. So its x-component is

- Vector B has a magnitude of 120 units and the direction is
(negative since it is below the x-axis), so the x-component is

So, vector A has the greater x component.
B) Vector B
Instead, the y-component of a vector can be found by using the formula

Here we have
- Vector B has a magnitude of 50 units along the positive x-direction, so
. So its y-component is

- Vector B has a magnitude of 120 units and the direction is
, so the y-component is

where the negative sign means the direction is along negative y:
So, vector B has the greater y component.