Math, asked by Anonymous, 4 months ago

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

Answers

Answered by Anonymous
2

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

Answered by iknowu9
1

Answer:

Step-by-step explanation:

Answer

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

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

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