tgA=3/4=BC/AC
Пусть BC=3X тогда AC=BC*4/3=3X*4/3=4X
по теореме пифогора AB^2 = AC^2 + BC^2 = 16*X^2 + 9*X^2 = 25*X^2 = (5*X)^2 ==>
AB=5X
sinA=BC/AB=3X/5X=3/5
