ТУ задача

[antoniy]

Screenshot_53.jpg (120.85 KiB) Прегледано 367 пъти

[ammornil]

$\triangle{ABC}, \quad AC\bot{BC}, \quad AB\in{\mu}, AC\notin{\mu}, BC\notin{\mu}, \\[6pt] \quad \angle{(BC,\mu)}=\beta, \angle{(AC,\mu)}=\alpha \\[6pt] \sin{\angle{\begin{pmatrix}p(ABC),\mu\end{pmatrix}}}=?\\[12pt]$

Screenshot 2025-05-04 230311.png (29.67 KiB) Прегледано 364 пъти

$\\[12pt]H\in{\mu}, CH\bot{\mu} \Rightarrow \begin{cases} \measuredangle{CAH}=\alpha \\ \measuredangle{CBH}=\beta \end{cases} \\[6pt] C_{1}\in{AB}, CC_{1}\bot{AB}, \quad \begin{cases} CH\bot{\mu} \Rightarrow CH\bot{AB} \\ CC_{1}\bot{AB} \end{cases}\hspace{0.5em} \text{Т-ма}3\bot: \Rightarrow HC_{1}\bot{AB} \Rightarrow \angle{HC_{1}C}=\angle{\begin{pmatrix}p(ABC),\mu\end{pmatrix}}=\varphi \\[6pt] CH=x \Rightarrow \begin{cases} \triangle{AHC}, \measuredangle{AHC}= 90^{\circ} \Rightarrow AC= \dfrac{x}{\sin{\alpha}} \\[12pt] \triangle{BHC}, \measuredangle{BHC}= 90^{\circ} \Rightarrow BC= \dfrac{x}{\sin{\beta}} \end{cases} \\[12pt] \triangle{ABC}: \quad AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{\dfrac{x^{2}}{\sin^{2}{\alpha}} + \dfrac{x^{2}}{\sin^{2}{\beta}}}= \dfrac{x}{\sin{\alpha}\sin{\beta}}\sqrt{\sin^{2}{\alpha}+\sin^{2}{\beta}} \\[12pt] S_{ABC}=\dfrac{AC\cdot{BC}}{2}=\dfrac{AB\cdot{CC_{1}}}{2} \Rightarrow CC_{1}=\dfrac{AC\cdot{BC}}{AB}= \dfrac{\dfrac{x}{\sin{\alpha}}\cdot{\dfrac{x}{\sin{\beta}}}}{\dfrac{x}{\sin{\alpha}\sin{\beta}}\sqrt{\sin^{2}{\alpha}+\sin^{2}{\beta}}} \\[6pt] \quad CC_{1}=\dfrac{x}{\sqrt{\sin^{2}{\alpha}+\sin^{2}{\beta}}} \\[12pt] \triangle{CHC_{1}}, \angle{C_{1}HC}=90^{\circ} \Rightarrow \sin{\varphi}=\dfrac{CH}{CC_{1}}=\dfrac{x}{\quad \dfrac{x}{\sqrt{\sin^{2}{\alpha}+\sin^{2}{\beta}}}\quad}=\sqrt{\sin^{2}{\alpha}+\sin^{2}{\beta}}$ $$\sin{\angle{\begin{pmatrix}p(ABC),\mu\end{pmatrix}}}= \sqrt{\sin^{2}{\alpha}+\sin^{2}{\beta}}$$