Prossze o rozwiazanie od 8 - 12 . Dziekuje slicznie  

z.8 I AC I² = 6² + 4² = 36 + 16 = 52  = 4*13 I AC I = √52 = √4*√13 = 2√13 Mamy zatem sin α = 4/IACI = 4/(2√13) = 2/√13 = (2√13)/13 cos α =  6/(AC I = 6/(2√13) = 3/√13 = (3√13)/13 tg α = 4/6 = 2/3 ctg α = 6/4 = 3/2 ---------------------- I BD I² = 6² - 4² = 36 - 16 = 20 = 4*5 I BD I = √20 = √4 *√5 = 2√5 mamy zatem sin β = 4/6 = 2/3 cos β = (2√5)/6 = √5/3 tg β = 4/(2√5) = 2/√5 = (2√5)/5 ctg β = (2√5)/4 = √5/2 =========================================================== z.9 a) A =(0;9), B = (3; 0) , C = ( 0; 2) AB = 3, AC = 2 BC = √(3² + 2²) = √13 sin α = 2/√13 = (2√13)/13 cos α = 3/√13 = (3√13)/13 tg α =  2/3 ctg α = 3/2 ------------ sin β = 3/√13 cos β = 2/√13 tg β = 3/2 ctg β = 2/3 =============== b) A = (0 ; 0), B = (6; 0), C = (6; 4) AB = 6 , Bc = 4 AC = √(6² + 4²) = √52 = √4*√13 = 2√13 sin α = 4/(2√13) = 2/√13 = (2√13)/13 cos α = 6/(2√13) = 3/√13 = (3√13)/13 tg α = 4/6 = 2/3 ctg α = 3/2 sin β = cos α = (3√13)/13 cos β = sin α = (2√13)/13 tg β = ctg α = 3/2 ctg β = tg α = 2/3 ====================================== c) A = (-6;1), B = (6 ; -4) , C = (6,1) AC = 6 - (-6) = 6 +6 = 12 BC = 0 - (-4) = 4 AB = √(12² + 4²) = √160 = √16 * √10 = 4√10 sin α = 4/(4√10) = 1/√10 = √10/10 cos α = 12/(4√10) = 3/√10 = (3√10)/10 tg α = 4/12 = 1/3 ctg α = 12/4 = 3 ------------------------ sin β = cos α = (3√10)/10 cos β = sin α = √10/10 tg β = ctg α = 3 ctg β = tg α = 1/3 ================================= d) A=(-1; -5), B =(4;5), C - (-4; 1) AB =√(4+1)²+ (5 +5)² = √25 =100 =√125 = 5√5 AC = √(-4+1)² +(1 +5)² = √(9 + 36) = √45 = √9*√5 = 3√5 BC = √(-4-4)² + (1 -5)² = √(64 + 16) = √80 = 4√5 sin α = AC/AB = 3√5 / 5√5 = 3/5 cos α = BC/AB = 4√5/5√5 = 4/5 tg α = AC / BC = 3√5/4√5 = 3/4 ctg α = 1/tg α = 4/3 ------------------------- sin β = 4/5 cos β = 3/5 tg β = 4/3 ctg β = 3/4 =============================================== cdn później