Question

номер 522 алгебра хелпаните плз

Answer 1

$\angle C = \dfrac{\pi}{2} = 90\textdegree$ , значит, $\bigtriangleup ABC$ прямоугольный. Сумма острых углов прямоугольного треугольника равна $90\textdegree$: $\angle A +\angle B = 90\textdegree$, откуда $\angle B = 90\textdegree - \angle A$.

а) $\angle A = \dfrac{\pi}{3} = 60\textdegree;\ \angle B = 90\textdegree - 60\textdegree = \boxed{\bf{30\textdegree}} = \boxed{\bf{\dfrac{\pi}{6}}}$

б) $\angle A = \dfrac{\pi}{4} = 45\textdegree;\ \angle B = 90\textdegree - 45\textdegree = \boxed{\bf{45\textdegree}} = \boxed{\bf{\dfrac{\pi}{4}}}$

в)  $\angle A = \dfrac{\pi}{5} = 36\textdegree;\ \angle B = 90\textdegree - 36\textdegree = \boxed{\bf{54\textdegree}} = \boxed{\bf{\dfrac{3\pi}{10}}}$

г)  $\angle A = \dfrac{\pi}{6} = 30\textdegree;\ \angle B = 90\textdegree - 30\textdegree = \boxed{\bf{60\textdegree}} = \boxed{\bf{\dfrac{\pi}{3}}}$

д)  $\angle A = \dfrac{2\pi}{9} = 40\textdegree;\ \angle B = 90\textdegree - 40\textdegree = \boxed{\bf{50\textdegree}} = \boxed{\bf{\dfrac{5\pi}{18}}}$

е)  $\angle A = \dfrac{\pi}{12} = 15\textdegree;\ \angle B = 90\textdegree - 15\textdegree = \boxed{\bf{75\textdegree}} = \boxed{\bf{\dfrac{5\pi}{12}}}$