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Hi everyone, I've been stuck on this relatively very simple problem for quite a while now and even after solving I'm still getting two answers while the book only lists one.

Question: If block D weighs 300 lb and block B weighs 275 lb, determine the required weight of block C and the angle (theta) for equilibrium.

So I did the usual since it's in equilibrium then F_{x}= 0 and F_{y}= 0.

So, F_{c}(cos30) - 275(cos[itex]\theta[/itex]) = 0

and F_{y}(sin30) + 275(sin[itex]\theta[/itex]) - 300 = 0

Then after finding cos[itex]\theta[/itex] in terms of F_{c}and substituting, you get this equation: 158.7cos[itex]\theta[/itex] + 275sin[itex]\theta[/itex] = 300

The answer listed is: [itex]\theta[/itex] = 40.9 and F_{c}= 240lb

Can anyone please tell me how to solve the equation above? I tried using the trig identity (square of cos plus square of sin = 1) but I got two answers, one was similar to the book ([itex]\theta[/itex] = 40.9, and one was [itex]\theta[/itex] 79). Thing is, when I use the second value I got (79) in the equations above, it works for all.

I'd appreciate some clarification on this problem! Thanks.

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# Homework Help: Simple Force in Equilibrium

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