∠BAC = ∠BAK - ∠CAK
tg∠BAK = 6 / 3,5 = 12/7
tg∠CAK = 6 / 6 = 1
tg∠BAC = tg(∠BAK - ∠CAK) =
= (tg∠BAK - tg∠CAK) / (1 + tg∠BAK ·· tg∠CAK) =
= (12/7 - 1) / (1 + 12/7) = (5/7) / (19/7) = 5/19
2.
g(9) = - f(9 - 12) / 2 = - f(- 3)/2 = f(3)/2 = 2/2 = 1
Так как функция f(x) - нечетная, то f(- 3) = - f(3)