What is the parallel-component of the weight

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To find the parallel component of weight for a 50 kg trunk on an 18-degree ramp, the correct formula is Fgx = W sin(θ). The weight (W) is calculated as 490 N (50 kg multiplied by 9.8 m/s²). The sine function must use degrees, so sin(18 degrees) is needed, not radians. After correcting the calculation, the answer aligns with option b, which is 47.6 N. Accurate unit conversion and angle measurement are crucial for solving such physics problems.
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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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