Solveeit Logo

Question

Question: Find the slope of the inclination of the line of the following:- \(\theta ={{30}^{\circ }}\). (a) ...

Find the slope of the inclination of the line of the following:- θ=30\theta ={{30}^{\circ }}.
(a) 3\sqrt{3}
(b) 13\dfrac{1}{\sqrt{3}}
(c) 23\dfrac{2}{\sqrt{3}}
(d) 32\dfrac{\sqrt{3}}{2}

Explanation

Solution

First understand the meaning of slope of a line and its geometrical significance. Now, to find the slope of the inclination of the given line, consider the given angle θ=30\theta ={{30}^{\circ }} as the inclination of the line with the positive direction of x – axis. Take the tangent function both the sides and use the trigonometric value tan30=13\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}} to get the answer.

Complete step-by-step solution:
Here we have been provided with the inclination of a line given as θ=30\theta ={{30}^{\circ }} and we are asked to find the slope of this inclination. First let us understand the meaning of slope of a line.
Now, slope of a line is the measure of change in the value of a function with respect to the change in the value of the variable on which the function depends. Mathematically it is given as the tangent value of the angle with which the line is inclined with the positive direction of x – axis. If we consider the inclination of the line with the positive x – axis as θ\theta then the slope is given as tanθ\tan \theta .
Let us come to the question, we have the inclination angle of the line given as θ=30\theta ={{30}^{\circ }}, so the slope will be given as: -
tanθ=tan30\Rightarrow \tan \theta =\tan {{30}^{\circ }}
We know that the value of tangent of the angle 30 degrees is 13\dfrac{1}{\sqrt{3}}, so we get,
tanθ=13\therefore \tan \theta =\dfrac{1}{\sqrt{3}}
Hence, option (b) is the correct answer.

Note: Note that if we are provided with two points on a straight line given as (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) then the value of the slope of the line is given as y2y1x2x1\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}. In case we have been provided with a curve instead of a straight line then the slope of this curve at a point (x, y) is the slope of the tangent line to this curve at this point. It is obtained by finding the value of the derivative dydx\dfrac{dy}{dx} at this point.