The figure drawn alongside (not drawn to scale) shows two straight lines AB and CD. If the equation of line AB is : x - $\sqrt{3}$y + 5 = 0 and the equation of line CD is : x - y = 2; write down the inclinations of lines AB and CD; also find the angle θ i.e. angle CPB.

Equation of line AB is :

⇒ x - $\sqrt{3}$y + 5 = 0

⇒ $\sqrt{3}$y = x + 5

⇒ y = $\dfrac{x}{\sqrt{3}} + \dfrac{5}{\sqrt{3}}$

Comparing above equation, with y = mx + c, we get :

m = $\dfrac{1}{\sqrt{3}}$

Let angle made by AB with positive side of x-axis be θ₁.

So, tan θ₁ = $\dfrac{1}{\sqrt{3}}$

⇒ tan θ₁ = tan 30°

⇒ θ₁ = 30°.

Equation of line CD is :

⇒ x - y = 2

⇒ y = x - 2

Comparing above equation, with y = mx + c, we get :

m = 1

Let angle made by CD with positive side of x-axis be θ₂.

So, tan θ₂ = 1

⇒ tan θ₂ = tan 45°

⇒ θ₂ = 45°.

From figure,

In △PWZ,

∠PZW = 180° - θ₂ (As, x axis is a straight line.)

∠WPZ = 180° - θ (As, AB is a straight line)

In △PWZ,

By angle sum property of triangle,

⇒ ∠PWZ + ∠PZW + ∠WPZ = 180°

⇒ θ₁ + 180° - θ₂ + 180° - θ = 180°

⇒ 30° + 180° - 45° + 180° - θ = 180°

⇒ 345° - θ = 180°

⇒ θ = 345° - 180° = 165°.

Hence, θ = 165°.