In an acute-angled triangle $ABC$, from the base $D$ of the altitude $BD$, perpendiculars $DM$ and $DN$ are dropped to the sides $AB$ and $BC$. It is known that $MN=a, BD=b$. Find the angle $ABC$.

#
Points $M$ and $N$ lie on a circle with diameter $B D$.

## Solution

Since $\angle B M D = \angle B N D = 90^{\circ}$, the segment $B D$ is seen from points $M$ and $N$ at a right angle, so points $M$ and $N$ lie on a circle with diameter $B D = 2 R$, where $R$ is the radius of this circle. Therefore,

$$
M N = 2 R \sin \angle A B C
$$

Thus,

$$
\sin \angle A B C = \frac{M N}{2 R} = \frac{M N}{B D} = \frac{a}{b}
$$