A force of 10 N acts along a line OB, which makes an angle of 60° with the horizontal line OA. What is the component of the force along the direction of the acute angle bisector of angle ∠AOB?
10 cos 30° N
This is a problem of resolving a force along a chosen direction. The component of a force along any line equals the force magnitude multiplied by the cosine of the angle between the force and that line: Fcomponent = F·cos θ, where θ is the angle between the force's line of action and the direction of interest.
Given: F = 10 N acts along OB, and OB makes 60° with the horizontal line OA, so ∠AOB = 60°.
Step 1 — Locate the bisector. The bisector of the acute angle ∠AOB splits the 60° angle into two equal halves, so it lies at 60°/2 = 30° from OA (and equally 30° from OB).
Step 2 — Angle between the force and the bisector. The force lies along OB (60° from OA) and the bisector lies at 30° from OA, so the angle between them = 60° − 30° = 30°.
Step 3 — Component along the bisector.
- Fbisector = F·cos(30°) = 10 cos 30° N ≈ 10 × 0.866 = 8.66 N.
Why the other choices are wrong: 10 cos 60° N would be the component along OA itself (the horizontal), since the force makes 60° with OA — not with the bisector. 10 sin 30° N uses the sine of the correct 30° angle, which gives the component perpendicular to the bisector, not along it. A value of 10 N (the full force) would only occur if the force were already aligned with the bisector, i.e., zero angle between them, which is not the case here.
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