[Trigonometric ratios in a right triangle] ] [ Rhombi. Criteria and properties $]$

In the rhombus $A B C D$, a perpendicular $B E$ is dropped from vertex $B$ to side $A D$. Find the angles of the rhombus if $2 \sqrt{3} C E=$ $\sqrt{7} A C$.
Apply the Law of Cosines to triangle $C A E$.

## Solution

Let $O$ be the point of intersection of the diagonals of the rhombus. Let $A C=2 x, \angle B A D=\alpha$. Then

$$
A B=\frac{A O}{\cos \angle B A O}=\frac{x}{\cos \frac{\alpha}{2}}, A E=A B \cos \angle B A E=\frac{x \cos \alpha}{\cos \frac{\alpha}{2}}, C E=\frac{A C \sqrt{7}}{2 \sqrt{3}}=x \sqrt{\frac{7}{3}}
$$

By the Law of Cosines in triangle $C A E$, we find that

$$
C E^{2}=A E^{2}+A C^{2}-2 A E \cdot A C \cos \angle C A E
$$

or

$$
\frac{7 x^{2}}{3}=\frac{x^{2} \cos ^{2} \alpha}{\cos ^{2} \frac{\alpha}{2}}+4 x^{2}-4 x^{2} \cos \alpha \frac{\cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}, \frac{7}{3}=4+\frac{\cos ^{2} \alpha}{\cos ^{2} \frac{\alpha}{2}}-4 \cos \alpha
$$

Using the formula $\cos ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2}$, after simplification, we obtain the equation

$$
6 \cos ^{2} \alpha+7 \cos \alpha-5=0
$$

from which we find that $\cos \alpha=\frac{1}{2}$. Therefore, $\alpha=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-29.jpg?height=377&width=757&top_left_y=1655&top_left_x=656)

| Problem 54004 Topics: | $\left[\begin{array}{ll}{\left[\begin{array}{l}\text { Constructions }\end{array}\right]} \\ {[\text { Angles between bisectors }]}\end{array}\right.$ | Difficulty: 4- |
| :---: | :---: | :---: |
|  | $[\quad$ Method GMT $]$ |  |

## Answer

$60^{\circ}, 120^{\circ}$