\[\boxed{\mathbf{911.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\overrightarrow{h} = k\overrightarrow{h}\]
\[\textbf{а)}\ 2 = |k| \bullet 0,5 \Longrightarrow |k| = 4;\]
\[\overrightarrow{m}\ \uparrow \downarrow \ \overrightarrow{h} \Longrightarrow k < 0 \Longrightarrow k = - 4.\]
\[\textbf{б)}\ 240 = |k| \bullet 12 \Longrightarrow |k| = 20;\]
\[\overrightarrow{m}\ \uparrow \uparrow \ \overrightarrow{h} \Longrightarrow k > 0 \Longrightarrow k = 20.\]
\[\textbf{а)}\ 400 = |k| \bullet 400 \Longrightarrow |k| = 1;\]
\[\overrightarrow{m}\ \uparrow \downarrow \ \overrightarrow{h} \Longrightarrow k < 0 \Longrightarrow k = - 1.\]
\[\textbf{а)}\ \sqrt{50} = |k| \bullet \sqrt{2} \Longrightarrow |k| =\]
\[= \sqrt{25} = 5;\]
\[\overrightarrow{m}\ \uparrow \uparrow \ \overrightarrow{h} \Longrightarrow k > 0 \Longrightarrow k = 5.\]
\[\boxed{\mathbf{911.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[BH\bot AC;\]
\[\angle ABE = \angle EBC.\]
\[\mathbf{Доказать:}\]
\[\angle OBE = \angle HBE.\]
\[\mathbf{Доказательство.}\]
\[1)\ Пусть\ \angle BCA = \alpha;\ \]
\[\angle ABE = \angle EBC = \beta:\]
\[\angle AOB = 2\alpha.\]
\[2)\ \mathrm{\Delta}ABO - равнобедренный:\]
\[\angle ABO = \frac{1}{2}(180{^\circ} - \angle AOB) =\]
\[= 90{^\circ} - \frac{1}{2}\angle AOB = 90{^\circ} - \alpha.\]
\[3)\ \angle OBE = \angle ABE - \angle ABO =\]
\[= \beta - (90{^\circ} - \alpha) = \alpha + \beta - 90{^\circ};\]
\[\angle HBE = \angle EBC - \angle HBC =\]
\[= \beta - (90{^\circ} - \alpha) = \alpha + \beta - 90{^\circ};\]
\[\angle OBE = \angle HBE.\]
\[Что\ и\ требовалось\ доказать.\]