Draw a right triangle with acute angle α at the left and another acute angle β at the upper right. Label the hypotenuse as x and the base as 4. Then the altitude is sqrt(x2 - 16).
Then x/4 = sec(α), so α = sec-1(x/4).
And 4/sqrt(x2 - 16) = tan(β), so β = tan-1( 4/sqrt(x2 - 16)).
What can you say about the angles α and β?