Trigonometric Equation: Solving for Theta with Quadratic Formula

[ThomasMagnus]

Solve for [tex]\theta[/tex]

20= 32[sin[tex]\theta[/tex]] (72/32[cos[tex]\theta[/tex]]) + 1/2(-9.8){(72/32[cos[tex]\theta[/tex]]}²I'm really having trouble with this question. I've tried it many times, but keep getting stuck.

Is there anyway to let X=72/32[cos[tex]\theta[/tex]] and then solve for x with the quadratic equation?

Can someone help me with this one? Thanks!

=)

 

[rl.bhat]

20= 32[sinLaTeX Code: \\theta ] (72/32[cosLaTeX Code: \\theta ]) + 1/2(-9.8){(72/32[cosLaTeX Code: \\theta ]}2

The above equation can be written as

20 = 72*tanθ + 1/2*(-9.8)(72/32)^2(sec^2θ)

20 = 72*tanθ + 1/2*(-9.8)(72/32)^2(1 + tan^2θ)

Now simplify the above equation and solve the quadratic to find tanθ.

 

[HallsofIvy]

Where did those fractions (tangent and secant) come from? I read it as
[tex]20= 72sin(\theta)cos(\theta)-(4.9)\left(\frac{9}{4}\right)^2cos^2(\theta)[/tex]

ThomasMagnus, if you do replace [itex]cos(\theta)[/itex] with "x", you will have to replace [itex]sin(\theta)[/itex] with [itex]\sqrt{1- x^2}[/itex] and you will eventually have a fourth degree equation, not quadratic. But that might be the only way to do it.

 

[Slats18]

So, the equation in question is

[tex]
20= 72sin(\theta)cos(\theta)-(4.9)\left(\frac{9}{4}\right)^2cos^2(\theta)
[/tex]

Would it help if we simplified the trig terms by way of double angle formulae?

For instance,

[tex]
2sin(\theta)cos(\theta)= sin(2\theta)
[/tex]

[tex]
2cos^2(\theta)-1= cos(2\theta)
[/tex]

could be used, which would make the equation look something like this:

[tex]
20=36sin(2\theta)-(9.8)\left(\frac{9}{4}\right)^2cos(2\theta)+1
[/tex]

Just throwing some things out there that could possibly help =)

 

[rl.bhat]

Where did those fractions (tangent and secant) come from? I read it as
[tex]20= 72sin(\theta)cos(\theta)-(4.9)\left(\frac{9}{4}\right)^2cos^2(\theta)[/tex]

ThomasMagnus, if you do replace [itex]cos(\theta)[/itex] with "x", you will have to replace [itex]sin(\theta)[/itex] with [itex]\sqrt{1- x^2}[/itex] and you will eventually have a fourth degree equation, not quadratic. But that might be the only way to do it.

y = vo*sin(θ)*t - 1/2*g(t^2)

And t = x/(vo*cosθ). Put it in the above equation.