The height of a right triangle, dropped to its hypotenuse, divides the bisector of an acute angle in the ratio $4: 3$, counting from the vertex.

Find the measure of this angle.
Let the height $C D$, drawn from the vertex $C$ of the right angle of the right triangle $A B C$, intersect the bisector $A K$ of angle $A$ at point $O$. Right triangles $D A O$ and $K A C$ are similar by two angles.

## Solution

Let the height $C D$, drawn from the vertex $C$ of the right angle of the right triangle $A B C$, intersect the bisector $A K$ of angle $A$ at point $O$. Right triangles $D A O$ and $C A K$ are similar by two angles, therefore, $\cos \angle A = \frac{A D}{A C} = \frac{A O}{A K} = \frac{4}{7}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-07.jpg?height=509&width=694&top_left_y=2404&top_left_x=682)

## Problem