How can I solve these two equations to find theta?

Question

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!

Answer 1

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

$$\tan{\theta} = \frac{5 \sqrt{3}}{3}$$

Answer 2

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 $\sin \theta$.

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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.

$\tan(\theta)= \dfrac{5}{\sqrt{3}}$

Answer 3

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 $$ t = \pm\frac{12}{7}\sqrt{7}.$$