ABCD rombda AC>BD va AC/BD-BD/AC=2 boʻlsa, ∠ A burchakni toping.

Shartga ko‘ra, rombning diagonallari $d_1$, $d_2$ tomoni $a$ va burchaklari $\angle A=\alpha$, $\angle B=180^\circ-\alpha$ bo‘lib, $d_1=AC$, $d_2=BD$ belgilashlarni kiritamiz. Bundan $\frac{AC}{BD}-\frac{BD}{AC}=2 \Leftrightarrow \frac{AC^2-BD^2}{AC\cdot BD}=2 \Leftrightarrow \frac{d_1^2-d_2^2}{d_1\cdot d_2}=2 \Leftrightarrow d_1^2-d_2^2=2d_1\cdot d_2$ (1) ga ega bo‘lamiz. Rombning yuzi uchun esa $S=\frac{d_1\cdot d_2}{2}=a^2\sin\alpha \Leftrightarrow d_1\cdot d_2=2a^2\sin\alpha$ tengliklarni hosil qilamiz. Ushbu $\triangle ABC$ da $\angle B=180^\circ-\alpha$, $\triangle BAD$ da $\angle A=\alpha$. Uchburchaklar uchun kosinuslar teoremasini qo‘llab $\begin{cases} d_1^2=2a^2+2a^2\cos\alpha \\ d_2^2=2a^2-2a^2\cos\alpha \end{cases}$ sistemani tuzib olamiz. Bu sistemani hadma-had ayirib, $d_1^2-d_2^2=4a^2\cos\alpha$ (2) ni hosil qilamiz. (1) va (2) tengliklardan sistema tuzamiz va undan $\alpha$ o‘tkir burchakni topamiz. $\begin{cases} d_1^2-d_2^2=4a^2\cos\alpha \\ d_1^2-d_2^2=2d_1\cdot d_2=4a^2\sin\alpha \end{cases} \Leftrightarrow 4a^2\cos\alpha=4a^2\sin\alpha$ $\cos\alpha=\sin\alpha$ $\mathrm{tg}\,\alpha=1$ bundan ($AC>BD$ shartga ko‘ra) $\alpha=45^\circ$ bo‘ladi.

Manba: UZBMB namunaviy test topshiriqlari (2021, mutaxassislik). A.A.Rahimqoriyev, M.A.Toxtaxodjayeva. Geometriya, umumiy o‘rta ta’lim maktablarining 8-sinfi uchun darslik. “O‘zbekiston” NMIU, Toshkent−2019