(i) Since \(\cos x = p\) and \(x\) is an acute angle, we use the identity \(\sin^2 x + \cos^2 x = 1\). Therefore, \(\sin^2 x = 1 - \cos^2 x = 1 - p^2\). Taking the positive square root (since \(x\) is acute), \(\sin x = \sqrt{1 - p^2}\).

(ii) \(\tan x = \frac{\sin x}{\cos x} = \frac{\sqrt{1 - p^2}}{p}\).

(iii) Using the identity \(\tan(90^\circ - x) = \cot x\), we have \(\tan(90^\circ - x) = \frac{1}{\tan x} = \frac{p}{\sqrt{1 - p^2}}\).