The horizontal and vertical components of a 100 N force inclined at 37° with the horizontal are: (consider the approximation of sin 37°=0.6, cos 37°=0.8 and tan 37°=0.75)
80 N and 60 N
Any force acting at an angle can be split into two mutually perpendicular parts — a horizontal component and a vertical component — using trigonometry. For a force F inclined at angle θ to the horizontal, the horizontal component is F·cos θ and the vertical component is F·sin θ. These two components, acting together, are exactly equivalent to the original force.
Given: F = 100 N, θ = 37°, with the standard approximations sin37° = 0.6, cos37° = 0.8, tan37° = 0.75.
Horizontal component:
- Fx = F·cos37° = 100 × 0.8 = 80 N
Vertical component:
- Fy = F·sin37° = 100 × 0.6 = 60 N
As a check, the components should recombine to the original magnitude: √(80² + 60²) = √(6400 + 3600) = √10000 = 100 N, confirming the result. Hence the components are 80 N (horizontal) and 60 N (vertical).
Why the other choices are wrong: 60 N and 80 N simply swaps the two values — it wrongly assigns F·sin θ to the horizontal direction and F·cos θ to the vertical, which would only be correct if the angle were measured from the vertical instead of the horizontal. 100 N and 0 N would apply only if the force were perfectly horizontal (θ = 0°), which contradicts the 37° inclination. 70 N and 70 N corresponds to a 45° force (equal components), not a 37° force.
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