5. In a right triangle $ABC$, a height $CH$ is drawn from the vertex of the right angle. From point $N$ on the leg $BC$, a perpendicular $NM$ is dropped to the hypotenuse, such that line $NA$ is perpendicular to $CM$ and $MH: CH=1: \sqrt{3}$. Find the acute angles of triangle $ABC$.
5. Solution. Since $\square A C B=90^{\circ}$ and $\square N M A=90^{\circ}$, a circle can be circumscribed around quadrilateral $A C N M$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d06b082d99fa6b9f8efdg-2.jpg?height=440&width=286&top_left_y=388&top_left_x=154)

Consider triangle $M C H$. It is a right triangle, so by the Pythagorean theorem we have:

$M C^{2}=M H^{2}+C H^{2}=M H^{2}+3 M H^{2}=4 M H^{2}$.

From this, we find $M H: M C=1: 2$, which implies that $\square M C H=30^{\circ}$. Since $M N \| C H$, then $\square N M C=\square M C H=30^{\circ}$. Angles NMC and NAC are inscribed and subtend the same arc, so $\square N A C=30^{\circ}$. We calculate $\square C M H: \square C M H=90^{\circ}-\square N M C=60^{\circ}$. From the perpendicularity of $M C$ and $A N$, it follows that $\square N A M=30^{\circ}$. Then $\square B A C=\square M A N+\square N A C=60^{\circ}$, and $\square A B C=30^{\circ}$.