tangent

$(x - 1 + 2)^2 = (2x - 4)^2 + (x - 1)^2$

$x^2 + 2x + 1 = 4x^2 - 16x + 16 + x^2 - 2x + 1$

$2x = 4x^2 - 18x + 16$

$x = 2x^2 - 9x + 8$

$2x^2 - 10x + 8 = 0$

$2(x^2 - 5x + 4) = 0$

2(x - 1)(x - 4) = 0

(x - 1)(x - 4) = 0

x = 4 

x = 1

However, side WU is (x - 1) so x can't be 1 since the side will be 0, which is impossible. 

Which means your answer is 4