In a right angled triangel the hypotenuse is four times as long as the perpendicular drawn to it from the opposite vertex .One of the acute angel is

c2 = h2 + (a-x)²

b2 = h2 + x²

on adding, we have  2h² + x² + (a-x)² = a² ..(1)

also a = 4h  and cos B = b/a = x/b

so x = b²/a

on solving (1), we have  h² + x² -4hx = 0

substituting above gives

b⁴ + a⁴/16 -a².b² =0 .. (2)

as cos B = b/a

on solving eqn. (2) we have ,  Cos²B = 0.933 or 0.067

Cos B = 0.966 or 0.259

B = 15 or 75 degrees

as B is acute , B is 15 degrees

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regards

Ramesh