\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[ABCD - квадрат;\]
\[\angle AKB = 74{^\circ}.\]
\[Найти:\]
\[\angle CAK.\]
\[Решение.\]
\[1)\ ABCD - квадрат:\]
\[\angle ACB = 90{^\circ};\]
\[AC - биссектриса\ \angle C;\]
\[\angle BCA = \frac{1}{2}\angle BCA = 45{^\circ}.\]
\[2)\ В\ \mathrm{\Delta}AKC:\]
\[\angle AKC = 180{^\circ} - \angle AKB = 106{^\circ};\]
\[\angle AKC + \angle ACK + \angle CAK = 180{^\circ}\]
\[106{^\circ} + 45{^\circ} + \angle CAK = 180{^\circ}\]
\[\angle CAK = 29{^\circ}.\]
\[Ответ:\ \ 29{^\circ}.\]