In a parallelogram, the length of one diagonal is e ( = 26 units), the angle between the diagonals is $\varphi\left(=65^{\circ}\right)$, and the difference between the sides is $d(=10$ units). What are the lengths of the sides and the acute angle of the parallelogram?
I. solution. In the parallelogram $ABCD$, let $AC=e, BA-BC=d$, and the magnitude relationship $BA>BC$ due to this, and $\varphi / e$ be. (Otherwise, independently of $\varphi$, for the formation of the triangle $ABC$, it is necessary that $d>0$ and $y>0$. According to this, $BA>BC$, and thus $BA-BC=d$, expressed in coordinates, and after squaring and rearranging:

$$
\begin{gathered}
\sqrt{\left(x+\frac{e}{2}\right)^{2}+y^{2}}=d+\sqrt{\left(x-\frac{e}{2}\right)^{2}+y^{2}} \\
2 e x-d^{2}=2 d \sqrt{\left(x-\frac{e}{2}\right)^{2}+y^{2}}
\end{gathered}
$$

After squaring and rearranging again,

$$
\frac{4 x^{2}}{d^{2}}-\frac{4 y^{2}}{e^{2}-d^{2}}=1
$$

Due to the squaring, $x$ and $y$, as well as $d$ and $e$, appear with even exponents, so if $B(x, y)$ satisfies this, then $D(-x,-y)$ also does, and their reflections over the axes do as well, for which $B^{\prime} A-B^{\prime} C=-d$. (The change of sign of $e$ gives the same with the exchange of $A$ and $C$.) This is essentially the equation of the hyperbola determined by the foci pair $A, C$ and the real axis length $d$.

The equation of the diagonal $BD$ is $y=x \cdot \operatorname{tg} \varphi$, from which (1) gives the coordinates of $B$ with trigonometric transformations and introducing the abbreviation $f$:

$$
x=\frac{d \sqrt{e^{2}-d^{2}} \cos \varphi}{2 \sqrt{e^{2} \cos ^{2} \varphi-d^{2}}}=\frac{f}{2} \cos \varphi(>0), \quad y=\frac{f}{2} \sin \varphi(>0)
$$

It is easy to see that here

$$
f=\frac{d \sqrt{e^{2}-d^{2}}}{\sqrt{e^{2} \cos ^{2} \varphi-d^{2}}}
$$

is precisely the length of the diagonal $BD$. A solution exists if $d<e \cos \varphi(<e)$.

Proceeding further, from the already written expression,

$$
BC=\sqrt{\left(x-\frac{e}{2}\right)^{2}+y^{2}}=\frac{1}{2} \sqrt{(f \cos \varphi-e)^{2}+(f \sin \varphi)^{2}}=\frac{1}{2} \sqrt{f^{2}+e^{2}-2 e f \cos \varphi}
$$

is identical to the expression obtained from the cosine theorem in the triangle $COB$. The other side is again $BA=d+BC$.

Finally, the angle of the parallelogram is obtained from the direction tangents of the rays $CB$ and $AB$:

$$
\operatorname{tg}\left(180^{\circ}-\gamma_{1}\right)=\frac{y}{x-\frac{e}{2}}, \quad \operatorname{tg} \gamma_{2}=\frac{y}{x+\frac{e}{2}}
$$

and from these, $\angle BCD=\gamma_{1}+\gamma_{2}$.

With the data of the problem, $f=52.70, \quad x=11.14, \quad y=23.88, \quad BC=23.95, \quad BA=33.95, \quad \angle BCD=85.55^{\circ}+44.69^{\circ}=130.24^{\circ}$, and the acute angle is $49.76^{\circ}=49^{\circ} 46'$.