# Suppose the angle formed by the line 14y= 23x and the positive x-axis is \theta. Find the tangent...

## Question:

Suppose the angle formed by the line 14y= 23x and the positive x-axis is {eq}\theta {/eq}. Find the tangent of {eq}\theta {/eq}.

## Slope of a Line:

(i) The slope of a line is the tangent of the angle formed by the line with the positive x-axis. i.e., if {eq}\theta {/eq} is the angle made by the line with the positive x-axis, then the slope of the line is, {eq}\tan \theta {/eq}.

(ii) The slope of the line {eq}y=mx+b {/eq} is {eq}m {/eq}.

(iii) From the above two points, we have {eq}m= \tan \theta {/eq}.

The given equation of the line is:

$$14y= 23x\\ \text{Dividing both sides by 14}, \\ y= \dfrac{23}{14}x$$

Comparing this with {eq}y=mx+b {/eq}, we get the slope to be {eq}m= \dfrac{23}{14} {/eq}.

The slope of the line is the tangent of the angle formed by the line with the positive x-axis. So we get:

$$\boxed{\mathbf{\tan \theta = \dfrac{23}{14}}}$$