So it says that a vector 140N is oriented at 35 deg with the horizontal. What is the magnitude of the horizontal component?
So I figured this is somewhat of a trig question with the vector being 140N and the theta or angle between this triangle being 35 degrees.

The horizontal component would be 140cos(35)

The vertical 140sin(35)

I agree with the horizontal component.

The vertical component would be 140sin(35).

DUH, that's what I meant. Mistyped. Thanks for the catch

I thought so.