Trigonometric equation

[ThomasMagnus]

Solve for \theta

20= 32[sin\theta] (72/32[cos\theta]) + 1/2(-9.8){(72/32[cos\theta]}2


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

ThomasMagnus, if you do replace cos(\theta) with "x", you will have to replace sin(\theta) with \sqrt{1- x^2} 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


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


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

For instance,


2sin(\theta)cos(\theta)= sin(2\theta)



2cos^2(\theta)-1= cos(2\theta)


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


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


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

ThomasMagnus, if you do replace cos(\theta) with "x", you will have to replace sin(\theta) with \sqrt{1- x^2} 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.