Question

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

Answer 1

ʜᴏʟᴀ ᴍᴀᴛᴇ !

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⭐ᴀɴsᴡᴇʀ⭐

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

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

Answer 2

Explanation:

parallel to slop=10cos45°=10*1/√2

=10/√2=5√2