(c) 60º
Since p is the length of perpendicular from origin on the straight line ax + by – p = 0.
p = \(\frac{|a.0+b.0-p|}{\sqrt{a^2+b^2}}\)
⇒ 1 = \(\sqrt{a^2+b^2}\) ⇒ 1 = a2 + \(\frac{3}{4}\) \(\big(\because{b} = \frac{\sqrt3}{2}\big)\)
⇒ a2 = \(\frac{1}{4}\) ⇒ a = \(\frac{1}{2}.\)
∴ Equation of the straight line is \(\frac{1}{2}x\) + \(\frac{\sqrt3}{2}y\) = p
⇒ \(x\) cos 60° + y sin 60° = p
Hence required angle is 60°, which is the angle between the perpendicular and the positive direction of x-axis.