Find the missing angle of similar triangle Find the missing angle $\theta$ in the triangle below given that $R>r$, $l\geq R$, $0< \theta < \frac{\pi}{2}$.

Attempted Solution
I attempted to use similar triangles to find the angle $\theta$ but the resulting expression for $\theta$ is pretty ugly.
$\frac{r}{R\cos\theta+\sqrt{l^2-R^2 \sin^2 \theta}}=\frac{R\sin\theta}{\sqrt{l^2-R^2\sin^2\theta}}$ 
I would then use some numerical solver to find $\theta$. Any other ways to attack this problem?


Hint
\begin{equation}
  \begin{cases}
    S = {S_1} + {S_2} = {1 \over 2}Rr\sin {\theta _1} + {1 \over 2}R{l_2}sin\theta  = {1 \over 2}r{l_2} \\ \\
    {\cos }{\theta _1}  = {{{r^2} + {R^2} - l_1^2} \over {2Rr}} \\ \\
    {\cos}\theta  = {{l_2^2 + {R^2} - {l^2}} \over {2R{l_2}}} \\ \\
    {\cos}{\theta _1} = {\sin}\theta , {\sin}{\theta _1} = {\cos}\theta \\ \\
    {r^2} + l_2^2 = {({l_1} + l)^2}
  \end{cases}
\end{equation}
We can find $l_1 ~ , l_2$ then we can find $\theta$
$S_1 \to $ The area of the triangle on the left
$S_2 \to $ The area of the triangle on the right

