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Mathematics

Find the inclination of the line whose gradient is

(i) 1

(ii) 3\sqrt{3}

(iii) 13\dfrac{1}{\sqrt{3}}

Straight Line Eq

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Answer

(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

3\sqrt{3} = tan θ
3\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

13\dfrac{1}{\sqrt{3}} = tan θ

13\dfrac{1}{\sqrt{3}} = tan 30°

∴ tan θ = tan 30°

∴ θ = 30°.

Hence, the inclination is 30°.

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