12) 1 .
AB =2*AH =2*(CH*ctqA) ==2*2*ctqA= 4ctqA ;
cosA =(√17)/17 =1/√17;
sinA =√(1 -cos²A) ²=√(1-1/17) =√16/17 =4/√17;
ctqA =cosA/sinA =1/4;
AB =4ctqA =1
13) 7/25 или [ 28/100 =0,28]
cosBAH =cos(90 -B) =sinB =sinBAC = 1/√(1+ctqBAC) =1/√(1+(24/7)²) =7/25
14) 4.
CH =AC*SinA =AC*1/√(1+ctq²A) =7* 1/√(1+(√33/4)²) =7*4/7 =4
tqA =4√33/33 =4/√33 ⇒ctqA =(√33)/4.
15)4/5 или [ 0,8]
cosB =cos(90° -A) =sinA ;
CH =√(AC² -AH)² =√(5² -3²) =4 .
sinA =CH/AC =4/5 .
cosB =sinA = 4/5.
(или сразу Пифагорова треугольник (3;4;5) )
********************* или
***sinA =√(1-cos²A) =√(1-(3/5)²) =4/5 ***