13) The inclination $$ heta $$ of the line with slope $$ =\sqrt{3} $$ is :

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13) The inclination $$ 	heta $$ of the line with slope $$ =\sqrt{3} $$ is :

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The inclination $$ \theta $$ is $$ 60^\circ $$. The slope $$ m $$ of a line is related to the angle of inclination $$ \theta $$ by the formula: $$ m = \tan(\theta) $$ Given that the slope $$ m = \sqrt{3} $$, we can set up the equation: $$ \tan(\theta) = \sqrt{3} $$ To find $$ \theta $$, we need to determine the angle whose tangent is $$ \sqrt{3} $$. We know that: $$ \tan(60^\circ) = \sqrt{3} $$ Thus, the angle of inclination $$ \theta $$ is: $$ \theta = 60^\circ $$ Therefore, the inclination $$ \theta $$ of the line with slope $$ \sqrt{3} $$ is: $$ \theta = 60^\circ $$

13) The inclination \( \theta \) of the line with slope \( =\sqrt{3} \) is :

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13) The inclination \( \theta \) of the line with slope \( =\sqrt{3} \) is :

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