Solve for Theta in bsin30-85.7sin(theta)=0

[Ry122]

in bsin30-85.7sin(theta)=0
How can I make theta the subject?

 

[Dick]

Solve for sin(theta) and use arcsin.

 

[Ry122]

how is that done?

 

[rocomath]

how is that done?

It's the inverse.

Recall ...

y=x^2,x\geq0

f^{-1}(x)=\sqrt x

So similarly ...

\sin x=y

\sin^{-1}y=x

 

[Ry122]

so is this correct
sin(theta)=-bsin30/-85.7
1/sin(theta)=(-85.7)/(-binsin30)sin^-1(-bsin30/-85.7)=(theta)

 

[Dick]

That doesn't make it the subject, now does it? If sin(theta)=a, then theta=arcsin(a). That's the definition of arcsin.

 

[rocomath]

I forgot to add that arcsine is defined in Quadrants I & IV. It's domain is: -\frac{\pi}{2}\leq y\leq\frac{\pi}{2}.

 

[rocomath]

so is this correct
sin(theta)=-bsin30/-85.7
1/sin(theta)=(-85.7)/(-binsin30)


sin^-1(-bsin30/-85.7)=(theta)

I think what you did was implied something like ... \frac{1}{x}=x^{-1}

It's not the same, the reciprocal of sine is cosecant. Refer to Dick's post.

 

[Dick]

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.

 

[Ry122]

That's it. arcsin(b*sin(30)/85.7)=theta.
How do you work this out on a calculator?

 

[Dick]

To work it out on a calculator you need to know b, right? The arcsin function is usually labeled sin^(-1).