Ответ:

Пошаговое объяснение:

7 . 1) ΔABC : AC = AB ctgα = a ctgα ;

2) ΔACD : x = AC sinβ = a ctgαsinβ ;

y = AC cosβ = a ctgαcosβ .

8 . 1) ΔABC : cosα = AC/AB ; AB = AC/cosα = b/cosα ;

2) ΔABD : BD = x = AB ctgβ = b ctgβ/cosα ;

sinβ = AB/AD ; AD = y = AB/sinβ = b/( cosαsinβ) .

9 . Проведемо BM⊥AD ; BC = MD = 6 ;

у ΔABM ∠ABM = 120° - 90° = 30° ; ∠ A = 60° ;

AM = BMctg60° = 2√3 * √3 = 2 * (√3 )² = 2 * 3 = 6 ;

AD = x = AM + MD = 6 + 6 = 12 .