Solve this system of trigonometric equations

Question

How to solve this system of equations analytically?

$$ \begin{cases} \:9\tan\alpha -\frac{4.9\cdot 9^2}{v^2\cos^2\alpha }=2.1\\ \:23\tan\alpha \:-\frac{4.9\cdot \:23^2}{v^2\cos^2\alpha \:}=2.44 \end{cases} $$

Answer 1

Let $x = v^2\cos^2\alpha$ and $y = \tan\alpha$.

For the first equation:

$$9y-\frac{4.9(9^2)}{x} = 2.1 \implies 9xy-396.9 = 2.1x \implies 9xy-2.1x = 396.9$$

$$\implies x(9y-2.1) = 396.9 \implies \color{blue}{x = \frac{396.9}{9y-2.1}}\tag{1}$$

For the second equation:

$$23y-\frac{4.9(23^2)}{x} = 2.44 \implies 23xy-2592.1 = 2.44x \implies 23xy-2.44u = 2592.1$$

$$x(23y-2.44) = 2592.1 \implies \color{green}{x = \frac{2592.1}{23y-2.44}}\tag{2}$$

So now, you have equations $(1)$ and $(2)$.

$$\color{blue}{x} = \color{green}{x}$$

$$\implies \frac{396.9}{9y-2.1} = \frac{2592.1}{23y-2.44}$$

Solve for $y$, obtain $\alpha$, solve for $x$, then plug $\alpha$ in $x = v^2\cos^2a$ to get $v$.