12.186. Let $O A$ be a fixed radius of a circle with center at point $O ; B$ be the midpoint of radius $O A ; M$ be an arbitrary point on the circle. Find the maximum value of angle $O M B$.
Solution.

Let $\angle O M B=\alpha, \angle O B M=\beta$ (Fig. 12.54).

From $\triangle M O B: \frac{\sin \beta}{\sin \alpha}=\frac{O M}{O B}=2 ; \sin \alpha=\frac{1}{2} \sin \beta$.

The maximum value of $\sin \beta$ is 1, therefore the maximum value of $\sin \alpha$ is $1 / 2$.

Since $O B<O M$, then $\alpha<\beta$, which means $\alpha$ is not the largest angle in triangle $M O B$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0937.jpg?height=615&width=583&top_left_y=76&top_left_x=96)

Fig. 12.55

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0937.jpg?height=496&width=506&top_left_y=196&top_left_x=720)

Fig. 12.56

Thus, $0<\alpha<\pi / 2$, and on the interval $(0 ; \pi / 2) y=\sin x$ is monotonically increasing, and the angle $\alpha$ takes its maximum value at the same point where $\sin \alpha$ takes its maximum value. Therefore, $\alpha=\pi / 6$.

Answer: $\pi / 6$.