How can I solve these two equations to find theta? I am doing a projectile motion questions and I have to solve these simultaneous equations:
$$\frac{-5t^{2}+30t\sin\theta }{30t\cos\theta }=\frac{1}{\sqrt3}$$
$$\frac{-10t+30\sin\theta }{30\cos\theta }=-\sqrt3$$
I solved them but the solution is long and tedious and I got theta = $-60^\circ$ which is wrong.
Thank you in advance!
 A: Hints:
$1.$ Cancel out $t$ in the first equation.
$2.$ Divide first equation by the second one, the $\cos \theta$ gets cancelled. And you're probably left with just $t$ and $\sin \theta$.

Alternate method:(Cancelling out $t$ throughout
Multiply the first equation by $2$.
$\dfrac{-10t+60 \sin(\theta)}{30 \cos \theta}=\dfrac{2}{\sqrt{3}}$
Second equation is :
$\dfrac{10t-30 \sin(\theta)}{30 \cos \theta}=\sqrt{3}$
Now add them both.
$\tan(\theta)= \dfrac{5}{\sqrt{3}}$
A: Subtracting the second from the first equation gives
$$\frac{t}{6\cos\theta} = \frac{4}{3}\sqrt{3}$$
or
$$t=8\sqrt{3}\cos\theta.$$
Plugging this back into the first equation gives
$$ \frac{-4\sqrt{3}\cos\theta+3\sin\theta}{3\cos\theta} = \frac{1}{3}\sqrt{3}$$
or
$$3\sin\theta = 5\sqrt{3}\cos\theta.$$
Thus finally we have
$$\tan\theta = \frac{5}{3}\sqrt{3}$$
and
$$ \frac{25}{3} = \tan(\theta)^2 = \cos(\theta)^{-2}-1,$$
thus
$$ t = \pm\frac{12}{7}\sqrt{7}.$$
A: Solve for $t$ in terms of $\theta$ in each equation.  
First:
$$t = 3 \sin{\theta} + 3 \sqrt{3} \cos{\theta}$$
Second:
$$t = 6 \sin{\theta} - 2 \sqrt{3} \cos{\theta}$$
Set the two versions of $t$ equatl to each other, and solve for $\theta$; I get
$$\tan{\theta} = \frac{5 \sqrt{3}}{3}$$
