Finding the angle of the force in degress

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The 4.9 N weight is in equilibrium under the influence of the three forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 5.9 N force acts from above on the right at an angle of 48◦ with the horizontal. The force 4.9 N acts straight down.

I found the magnitude  of force F its 39.58

please help me find the angle α of the force F
Answer in units of ◦

Jun 2nd, 2015

Thank you for the opportunity to help you with your question!

for equillibrium, 39.58*sin (alpha)+5.9*sin48=4.9

39.58*sin (alpha)+4.38455=4.9

39.58*sin (alpha)=4.9-4.38455

sin alpha=0.51545*39.58

sin alpha=20.40

alpha= sin^(-1) 20.40

alpha=

i think the force F is incorrect, let me find it first

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 3rd, 2015

hello, 

F=sqrt(4.9^2+3.948^2)

F=sqrt(39.595)

F=6.29 N.

so from above procedure, 

using sine rule, 

sin alpha= 4.9/6.29

sin alpha=0.7790

alpha=51.17.

please ignore the previous answer

Jun 3rd, 2015

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