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Answer:
u= 2 ; v = 1
Sen 60 = [tex]\frac{\sqrt{3}}{u}[/tex] so u=2
Using the pythagoras theorem [tex]u^{2} =v^{2} +\sqrt{3}^{2}[/tex] so v =[tex]\sqrt{2^{2} - 3}[/tex] =1