(a) 60º
The 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 = \(\big|\frac{m_1-m_2}{1+m_1m_2}\big|\) = \(\bigg|\frac{(2-\sqrt3)-(2+\sqrt3)}{1+(2-\sqrt3)(2+\sqrt3)}\bigg|\)
= \(\bigg|\frac{-2\sqrt3}{1+1}\bigg|=\sqrt3\)
∴ θ = –1 tan ( 3) = 60º.