Question

Resolve a weight of 10 N in a direction parallel to a slope inclined at 45° to the horizontal.

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Answer 1

Answer:

Let the force 10N makes an angle Φ with the inclined line

• Parallel component = 10 × sinΦ N ;
• Perpendicular component = 10 × cosΦ N

Answer 2

$\fbox\red{Answer}$

Let the force 10N makes an angle θ with the inclined line

parallel component = 10×sinθ N ; perpendicular component = 10×cosθ N