What is the parallel-component of the weight

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The discussion focuses on calculating the parallel component of weight for a trunk resting on a ramp inclined at 18 degrees. The correct formula to use is Fgx = W sin(θ), where W is the weight of the trunk. The weight of the trunk is 490 N, leading to the calculation Fgx = 490 N * sin(18°). The correct answer is 151 N, corresponding to option c. The confusion arose from using radians instead of degrees in the sine function.

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Jess048
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A 50 kg trunk rest on a ramp at 18 degrees. What is the parallel-component of the weight?
a. 15.5 N
b. 47.6 N
c. 151 N
d. 466 N

So far I got:
I used the formula Fgx = W sin 0.

Fgx= -(490 N) sin(18)
My answer was 368, as you see it is not one of the choices. Am i leaving out a step or am I not finished with the problem. Can someone clarify please thanx.
 
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You are computing the sine of 18 radians, not 18 degrees.
 
Oh ok so the answer would be C
 

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