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in bsin30-85.7sin(theta)=0
How can I make theta the subject?
How can I make theta the subject?
It's the inverse.how is that done?
I think what you did was implied something like ... [tex]\frac{1}{x}=x^{-1}[/tex]so is this correct
sin(theta)=-bsin30/-85.7
1/sin(theta)=(-85.7)/(-binsin30)
sin^-1(-bsin30/-85.7)=(theta)
Yes. That's it. arcsin(b*sin(30)/85.7)=theta.so is this correct
sin(theta)=-bsin30/-85.7
1/sin(theta)=(-85.7)/(-binsin30)
sin^-1(-bsin30/-85.7)=(theta)