Answer
$\theta=60.74^o$
Work Step by Step
The figure from the problem was redrawn below. As the pebble is released from the circle from point P, it will travel along tangent line PT. In other words, we have $\angle TPC=90^o$
We have $d=10r$. According to law of sine, $$\frac{\sin \angle T}{\sin \angle P}=\frac{r}{d}=\frac{1}{10}$$ $$\sin\angle T=\frac{1}{10}\sin90=0.1$$ $$\angle T=5.74^o$$
We both have $$35^o+\angle TCP+\theta=\angle P+\angle T+\angle TCP=180^o$$ $$35^o+\theta=\angle P+\angle T=90^o+5.74^o=95.74^o$$ $$\theta=60.74^o$$
