[The triangle formed by the bases of two altitudes and a vertex]

The side of the triangle is $\sqrt{2}$, and the angles adjacent to it are $75^{\circ}$ and $60^{\circ}$.

Find the segment connecting the bases of the altitudes drawn from the vertices of these angles.
Let $B M$ and $C N$ be the altitudes of triangle $A B C$, $B C=\sqrt{2}$, $\angle B=75^{\circ}$, $\angle C=60^{\circ}$. Then $\angle A=45^{\circ}$.

Triangle $A M N$ is similar to triangle $A B C$ with a similarity coefficient of $\cos 45^{\circ}$. Therefore, $M N=$ $B C \cdot \cos 45^{\circ}=1$.

## Answer

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