If the hypotenuse of right-angled triangle is four times the length of perpendicular drawn from opposite vertex to it, then the difference of two acute angles will be:

60°

$Δ=\frac{1}{2}ab=\frac{1}{2}P.4P⇒ab=4P^2$

also $a^2+b^2+c^2=16p^2⇒(a-b)^2=a^2+b^2-2ab=16p^2-8p^2=8p^2$

and $(a+b)^2=24p^2;\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac{c}{2}=\frac{1}{\sqrt{3}}.1$

$∴\frac{A-B}{2}=30°⇒A-B=60°$