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Sagot :
To produce work, force must be parallel to distance therefore, the answer is d.
Let me do a proof to let you remember quickly:
[tex]sin\theta = \frac{x}{hypotenuse}[/tex]
We know from trigonometric identity of tangent function is
[tex]tan\theta = \frac{sin\theta}{cos\theta} [/tex]
To let you remember quickly, let me prove it to you that:
[tex]tan\theta= \frac{sin\theta}{cos\theta}= \frac{y}{x} [/tex]
(Please be guided by the picture I uploaded)
[tex]tan\theta=\frac{sin\theta}{cos\theta}[\tex] ; [tex] cos\theta= \frac{adjacent}{hypotenuse}[/tex] ; [tex]sin\theta= \frac{opposite}{hypotenuse} [/tex] ; [tex]opposite = y[/tex] ; [tex]adjacent=x[/tex]
[tex]\therefore tan\theta=\frac{ \frac{y}{hypotenuse} }{ \frac{x}{hypotenuse} } [/tex]
[tex]tan\theta= \frac{y(hypotenuse)}{x(hypotenuse)}[/tex]
[tex]\therefore tan\theta= \frac{y}{x} [/tex]
Let me do a proof to let you remember quickly:
[tex]sin\theta = \frac{x}{hypotenuse}[/tex]
We know from trigonometric identity of tangent function is
[tex]tan\theta = \frac{sin\theta}{cos\theta} [/tex]
To let you remember quickly, let me prove it to you that:
[tex]tan\theta= \frac{sin\theta}{cos\theta}= \frac{y}{x} [/tex]
(Please be guided by the picture I uploaded)
[tex]tan\theta=\frac{sin\theta}{cos\theta}[\tex] ; [tex] cos\theta= \frac{adjacent}{hypotenuse}[/tex] ; [tex]sin\theta= \frac{opposite}{hypotenuse} [/tex] ; [tex]opposite = y[/tex] ; [tex]adjacent=x[/tex]
[tex]\therefore tan\theta=\frac{ \frac{y}{hypotenuse} }{ \frac{x}{hypotenuse} } [/tex]
[tex]tan\theta= \frac{y(hypotenuse)}{x(hypotenuse)}[/tex]
[tex]\therefore tan\theta= \frac{y}{x} [/tex]

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