Solve a system of trigonometric equations

Question

How can I solve this system of trigonometric equations analytically? It is from physics class. $$ \begin{cases} 30t\cos{\alpha}=50\\ -30t\sin{\alpha}-4.9t^2=0 \end{cases} $$

Answer 1

Hint: Squaring both the equations, you will get $900t^2\cos^2{\alpha}=2500\\ 900t^2\sin^2{\alpha}={4.9}^2t^4$.

Note that $\sin^2{\alpha}+\cos^2{\alpha}=1$.

So add both the equations and solve for $t$ using the substitution $t^2=u$.

Answer 2

$30t\cos{\alpha}=50 \implies t=\frac{5}{3} \sec\alpha$

You can plug this information into the other equation and solve:

$$-30t\sin{\alpha}-4.9t^2=0\implies -30(\frac{5}{3} \sec\alpha)\sin{\alpha}-4.9(\frac{5}{3} \sec\alpha)^2=0$$

$$-50(\tan\alpha)-4.9(\frac{5}{3} \sec\alpha)^2=0$$

$$-50(\tan\alpha)-4.9\frac{25}{9} \sec^2\alpha=0$$

$$-50(\tan\alpha)-4.9\frac{25}{9} (\tan^2\alpha+1)=0$$

Taking $y=\tan\alpha$ you can solve a quadratic equation.
$$-50(y)-4.9\frac{25}{9} (y^2+1)=0$$

I think you're probably in good shape from here?

Answer 3

Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.

This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$