a = i + 5j and b = 5i - j

cos\(\theta\) = \(\frac{a.b}{|a|.|b|}\)

= \(\frac{(1 \times 5) + (5x - 1)}{(\sqrt{1^2 + 5^2}) (5^2 + (-1))^2}\)

= \(\frac{5 - 5}{\sqrt{26}\times \sqrt{26}}\) = 0

x = cos\(^{-1}\)(0), x = 90\(^o\)