projectile motion find the angle

Question

I would please like help on the following question related to projectile motion.

An object is projected in a $60$ m/s velocity with a $\theta$ angle to the ground. If the object has $30^\circ$ angle to the ground at a height of $120$ m from the projected point, find $\theta$.

I seperated the velocity at above position into to parts and found the velocities for each factors seperately.

↑ v = u+at
    = u sin θ - 10t

→v = u+at
   =u cos θ

tan 30 = u sin θ - 10t / u cos θ

What to do after this result ?

Answer 1

\begin{align*} y &= 60t\sin \theta-5t^2 \\ 120 &= 60t\sin \theta-5t^2 \\ 24 &= t(12\sin \theta-t) \quad \cdots \cdots (1) \\ \frac{v_{y}}{v_{x}} &= \frac{60\sin \theta-10t}{60\cos \theta} \\ \frac{1}{\sqrt{3}} &= \frac{6\sin \theta-t}{6\cos \theta} \\ t &= 6\sin \theta-2\sqrt{3} \cos \theta \quad \cdots \cdots (2) \\ \end{align*}

Let $s = \sin \theta \, $ and substitute $(2)$ into $(1)$, \begin{align*} 24 &= \left[ 6s-2\sqrt{3(1-s^2)} \, \right] \left[ 6s+2\sqrt{3(1-s^2)} \, \right] \\ 24 &= 36s^2-12(1-s^2) \\ 4s^2 &= 3 \\ \sin \theta &= \frac{\sqrt{3}}{2} \\ \theta &= 60^{\circ} \end{align*}