## Trigonometry (11th Edition) Clone

As $v$ increases without bound, the equation of the asymptote is $\theta = 45^{\circ}$
Let $~~h=6$ $\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+64h}}\right)$ $\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+(64)(6)}}\right)$ $\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+384}}\right)$ As $v$ increases, the value of ${\frac{v^2}{2v^2+384}}$ approaches $\frac{1}{2}$. We can find the equation of the asymptote: $\theta = arcsin \left(\sqrt{\frac{1}{2}}\right)$ $\theta = arcsin \left(0.707\right)$ $\theta = 45^{\circ}$ As $v$ increases without bound, the equation of the asymptote is $\theta = 45^{\circ}$