Find the inclination of the line whose gradient is

(i) 1

(ii) $\sqrt{3}$

(iii) $\dfrac{1}{\sqrt{3}}$

(i) Let inclination be θ.

We know that,

m = slope or gradient = tan θ

⇒ 1 = tan θ
⇒ 1 = tan 45°
∴ tan θ = tan 45°
∴ θ = 45°.

Hence, the inclination is 45°.

(ii) Let inclination be θ.

We know that,

m = slope or gradient = tan θ

⇒ $\sqrt{3}$ = tan θ
⇒ $\sqrt{3}$ = tan 60°
∴ tan θ = tan 60°
∴ θ = 60°

Hence, the inclination is 60°.

(iii) Let inclination be θ.

We know that,

m = slope or gradient = tan θ

⇒ $\dfrac{1}{\sqrt{3}}$ = tan θ

⇒ $\dfrac{1}{\sqrt{3}}$ = tan 30°

∴ tan θ = tan 30°

∴ θ = 30°.

Hence, the inclination is 30°.