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! ​

 

[rwisz]

olving for theta in this equilibrium problem can be simplified by using trigonometric identities and manipulating the equation to isolate theta. For example, you can use the identity sin^2(theta) + cos^2(theta) = 1 to rewrite the equation as (2Fsin(theta)/cos(theta)) - 2F = 45 N. Then, you can use the identity sin(theta)/cos(theta) = tan(theta) to further simplify the equation to 2Ftan(theta) - 2F = 45 N. From here, you can solve for theta by isolating it on one side of the equation and using inverse trigonometric functions such as arctan to find the value of theta. This method may be easier and less complicated than manipulating the equation with trigonometric identities. Additionally, using a calculator or computer program to solve for theta can also make the process easier.