One of the sides of the parallelogram is equal to 10, and the diagonals are equal to 20 and 24. Find the cosine of the acute angle between the diagonals.

#
Let the diagonals $A C$ and $B D$ of parallelogram $A B C D$ intersect at point $O$. Then triangle $A B O$ is isosceles.

## Solution

Let the diagonals $A C$ and $B D$ of parallelogram $A B C D$ intersect at point $O$, where $A C=20, B D=24, A B=C D=10$. Since the diagonals of a parallelogram are bisected by their point of intersection, we have $A O=10, B O=12$. Therefore, triangle $A B O$ is isosceles. Let $A H$ be its height. Then $O H=\frac{1}{2} B O=6$. From the right triangle $A O H$, we find that

$$
\cos \angle A O B=\frac{O H}{A O}=\frac{6}{10}=\frac{3}{5}
$$

Let the diagonals $A C$ and $B D$ of parallelogram $A B C D$ intersect at point $O$, where $A C=20, B D=24, A B=C D=10$. Since the diagonals of a parallelogram are bisected by their point of intersection, we have $A O=10, B O=12$. Therefore, triangle $A B O$ is isosceles. Let $A H$ be its height. Then $O H=\frac{1}{2} B O=6$. From the right triangle $A O H$, we find that

$$
\cos \angle A O B=\frac{O H}{A O}=\frac{6}{10}=\frac{3}{5}
$$

Let the diagonals $A C$ and $B D$ of parallelogram $A B C D$ intersect at point $O$, where $A C=20, B D=24, A B=C D=10$. Since the diagonals of a parallelogram are bisected by their point of intersection, we have $A O=10, B O=12$. Therefore, triangle $A B O$ is isosceles. Let $A H$ be its height. Then $O H=\frac{1}{2} B O=6$. From the right triangle $A O H$, we find that

$$
\cos \angle A O B=\frac{O H}{A O}=\frac{6}{10}=\frac{3}{5}
$$

## Answer