Let's let $\theta$ be the angle of elevation. Let's orient our coordinate axes such that the vertical axis $y$ points up in the positive direction, and the horizontal axis $x$ points in the direction of motion in the positive direction.

For the vertical component of motion, we have:

$ \displaystyle a_y=-g$

$ \displaystyle v_y=-gt+v_0\sin(\theta)$

$ \displaystyle y=-\frac{g}{2}t^2+v_0\sin(\theta)t$

For the horizontal component of motion, we have:

$ \displaystyle a_x=0$

$ \displaystyle v_x=v_0\cos(\theta)$

$ \displaystyle x=v_0\cos(\theta)t$

Let's eliminate the parameter $t$, by using $ \displaystyle t=\frac{x}{v_0\cos(\theta)}$:

And so:

$ \displaystyle y=-\frac{g}{2}\left(\frac{x}{v_0\cos(\theta)}\right)^2+v_0\sin(\theta)\left(\frac{x}{v_0\cos(\theta)}\right)=-\frac{g}{2v_0^2\cos^2(\theta)}x^2+\tan(\theta)x$

Multiply through by $\cos^2(\theta)$:

$ \displaystyle y\cos^2(\theta)=-\frac{g}{2v_0^2}x^2+\sin(\theta)\cos(\theta)x$

Using double-angle identities, and multiplying through by 2, we may write:

$ \displaystyle y\left(\cos(2\theta)+1\right)=-\frac{g}{v_0^2}x^2+\sin(2\theta)x$

Arrange as:

$ \displaystyle y+\frac{g}{v_0^2}x^2=\sin(2\theta)x-\cos(2\theta)y$

Using a linear combination identity, we have:

$ \displaystyle \sqrt{x^2+y^2}\sin\left(2\theta-\arctan\left(\frac{y}{x}\right)\right)=y+\frac{g}{v_0^2}x^2$

Solving for $\theta$, there results:

$ \displaystyle \theta=\frac{1}{2}\left(\arcsin\left(\frac{v_0^2y+gx^2}{v_0^2\sqrt{x^2+y^2}}\right)+\arctan\left(\frac{y}{x}\right)\right)$

Now, all that's left is to plug in the given values.