Problem: Let $ABC$ be an acute-angled triangle with $O$ as its circumcenter.
Let $P$ on line $BC$ be the foot of the altitude from $A$.
Assume that $\angle BCA \ge \angle ABC + 30\dg$.
Prove that $\angle CAB + \angle COP < 90\dg$.

Solution: The conclusion rewrites as
\begin{align*}
  \angle COP &< 90\dg - \angle A = \angle OCP \\
  \iff PC &< PO \\
  \iff PC^2 &< PO^2 \\
  \iff PC^2 &< R^2 - PB \cdot PC \\
  \iff PC \cdot BC &< R^2 \\
  \iff ab \cos C  &< R^2 \\
  \iff \sin A \sin B \cos C & < \frac14.
\end{align*}
Now
\[ \cos C \sin B
  = \half \left( \sin(C+B)-\sin(C-B) \right)
  \le \half \left( 1 - \half \right) = \frac 14 \]
which finishes when combined with $\sin A < 1$.

\begin{remark*}
  If we allow $ABC$ to be right then
  equality holds when $\angle A = 90\dg$,
  $\angle C = 60\dg$, $\angle B = 30\dg$.
  This motivates the choice of estimates
  after reducing to a trig inequality.
\end{remark*}