cos(b)=BH/BC => BC=BH/cos(b) => BC=4/cos(b)
С другой стороны
cos(x)=BH/AB => AB=BH/cos(x) = > AB=4/cos(x)
По теореме Пифагора
(AC)^2=(AB)^2+(BC)^2=16/(cos(b)^2+16/(cos(x)^2)
AC=4*sqrt(1/(cos(b)^2+cos(x)^2)
cos(b)=BH/BC => BC=BH/cos(b) => BC=4/cos(b)
С другой стороны
cos(x)=BH/AB => AB=BH/cos(x) = > AB=4/cos(x)
По теореме Пифагора
(AC)^2=(AB)^2+(BC)^2=16/(cos(b)^2+16/(cos(x)^2)
AC=4*sqrt(1/(cos(b)^2+cos(x)^2)