10.342. The lengths of the sides and diagonals of a parallelogram are equal to $a, b, c$ and $f$ respectively. Find the angles of the parallelogram if $a^{4}+b^{4}=c^{2} f^{2}$.
Solution.

Let $a$ and $\beta$ be the angles of the given parallelogram, opposite to the diagonals $c$ and $f$ respectively, and $0^{\circ}<\alpha<90^{\circ}$. Then $c^{2}=a^{2}+b^{2}-2 a b \cos \alpha$; $f^{2}=a^{2}+b^{2}-2 a b \cos \beta=a^{2}+b^{2}+2 a b \cos \alpha$. Multiplying the equations of the system $\left\{\begin{array}{l}c^{2}=a^{2}+b^{2}-2 a b \cos \alpha ; \\ f^{2}=a^{2}+b^{2}+2 a b \cos \alpha,\end{array}\right.$ we get: $f^{2} c^{2}=\left(a^{2}+b^{2}\right)^{2}-4 a^{2} b^{2} \cos ^{2} \alpha$. Then $a^{4}+b^{4}=a^{4}+2 a^{2} b^{2}+b^{4}-4 a^{2} b^{2} \cos ^{2} \alpha ; \cos ^{2} \alpha=\frac{1}{2} ; \cos \alpha=\frac{1}{\sqrt{2}}$; $\alpha=45^{\circ} ; \beta=180-\alpha=135^{\circ}$.

Answer: $45^{\circ} ; 135^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0804.jpg?height=410&width=656&top_left_y=869&top_left_x=350)

Fig. 10.126