(a) 60ºThe two lines are: y = \((2-\sqrt3)\) \(x\) + 5 ...(i) y = \((2-\sqrt3)\) \(x\) – 7 ...(ii) Slope of line (i), m1 = \(2-\sqrt3\)Slope of line (ii), m2 = \(2+\sqrt3\)If θ is the angle between the two lines, then tan θ = \(\big|rac{m_1-m_2}{1+m_1m_2}\big|\) = \(\bigg|rac{(2-\sqrt3)-(2+\sqrt3)}{1+(2-\sqrt3)(2+\sqrt3)}\bigg|\)= \(\bigg|rac{-2\sqrt3}{1+1}\bigg|=\sqrt3\)∴ θ = –1 tan ( 3) = 60º.