Is there an easier way to solve for theta in this equilibrium problem?

[quantum_enhan]

Homework Statement

The Attempt at a Solution

2Fsin(theta)=45 N

Where F=ks=(75N/m)(s-1.2m).

Looking at the triangle, we can state that s, the stretched length is equal to 1.2m/cos(theta). Therefore,

2(75N/m)[(1.2m/cos(theta))-1.2m]sin(theta)=45 N
sin(theta)[(1/cos(theta))-1]=0.25

From here, the math gets tricky and complicated for me. Is there any easier way to solve theta?:S

 

[tiny-tim]

hi quantum_enhan!

(have a theta: θ )

sin(theta)[(1/cos(theta))-1]=0.25

From here, the math gets tricky and complicated for me. Is there any easier way to solve theta?:S

You have sinθ(secθ - 1) = 0.25.

Try substituting t = tan(θ/2).

 

[quantum_enhan]

I don't see how you can get tan(θ/2) from sinθ(secθ - 1). Don't you need (1-cosθ)/sinθ. I can't seem to get it to equal the half angle identity..

 

[tiny-tim]

You can get sinθ from tan(θ/2), and you can get cosθ from tan(θ/2) …

so just build it up! ​