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Resolving forces angle Q

Haven't attached the Q as all it is is the forces diagram drawn different. My question isn't about what they did - I get that. Initially I wanted to find the angle of where the the R arrow and 2N arrow point together. To do this, I did sin5/R = sin angle/5 and then solved for the angle. But I keep getting 8. something and this doesn't work in the context???
(edited 2 years ago)
Screenshot 2021-10-14 161148.png105.0KB
Original post by velaris08
Haven't attached the Q as all it is is the forces diagram drawn different. My question isn't about what they did - I get that. Initially I wanted to find the angle of where the the R arrow and 2N arrow point together. To do this, I did sin5/R = sin angle/5 and then solved for the angle. But I keep getting 8. something and this doesn't work in the context???


So if sin(x) = 0.14465, x = 8.317 deg or .... another angle.
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velaris08
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Original post by old_engineer
So if sin(x) = 0.14465, x = 8.317 deg or .... another angle.


Yes - but since we find the other angle (not 5 or 8.317) to be 53.316 - how does that make sense?
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mqb2766
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Original post by velaris08
Yes - but since we find the other angle (not 5 or 8.317) to be 53.316 - how does that make sense?

There is another solution to asin(0.14465) as stated by oldengineer.

Just draw the triangle properly to get some insight. The side lengths are 2,3.01,5. As the angle is opposite the 5 side, it must be the largest angle in the triangle.
Reply 4
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velaris08
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Original post by mqb2766
There is another solution to asin(0.14465) as stated by oldengineer.

Just draw the triangle properly to get some insight. The side lengths are 2,3.01,5. As the angle is opposite the 5 side, it must be the largest angle in the triangle.


Ohhhhh I forgot about the thing where you can have two different angle sizes with sin. But how does it work here? When I draw it out it seems like everything is locked in place?
(edited 2 years ago)
Reply 5
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mqb2766
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Original post by velaris08
Ohhhhh I forgot about the thing where you can have two different angle sizes with sin. But how does it work here? When I draw it out it seems like everything is locked in place?

Not sure what you mean by locked. You can have one obtuse and one acute triangle with the information used in the sin rule. The extra 2 length, is the tie breaker here to determine which is which. Upload your attempted sketches if youre still stuck.

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