Question

Angle of line with +ve direction of x-axis is $\theta$. The line is rotated about some point in it in anticlockwise direction by angle ${45^{^0}}$and its slope become $3\;,Find\;\theta .$

Answer 1

We should have knowledge of slope of a straight line & some basic knowledge of trigonometry. Formula of slope of a straight line should be applied & solved to get the angle between that line & positive x axis.

Complete step-by-step answer:
Given, the angle of line with the +ve direction of the x-axis is $\theta$.
We know , slope or gradient is defined as a number that describes both the direction & steepness of the line. The slope is represented by $\operatorname{m} = tan\theta$, where m is the slope of that line which forms $\theta$ angle with the x-axis .
When a line is rotated anticlockwise from +ve x axis, it forms an angle in the 1st quadrant.
As per question , let's suppose the line AB is rotated about point A in anticlockwise direction by angle ${45^0}$ and become AC.
Now, applying $\operatorname{m} = tan\theta$ by the definition of slope $forAC$
$\operatorname{Slope} \left( m \right) = tan\theta = 3$
$\therefore tan\left( {45 + \theta } \right) = 3$
$\Rightarrow 45 + \theta = 71.565$ [ taking inverse of tan as we know if $\tan \theta = x$ then $\theta = {\tan ^{ - 1}}x$ by using calculator, you can get value of inverse of tan for given value]
$\Rightarrow \theta = 71.565 - 45$
$\Rightarrow 45 + \theta = 71.565$ [ solving for $\theta$ ]
$\Rightarrow \theta = 26.565 = 26.5{7^0}\left( {approximately} \right)$
Hence those lines make an angle of $26.57^\circ$ with the +ve x-axis.

Note: In this type of problem, the angle of the straight line varies when the line is rotated clockwise or anticlockwise in any direction.The slope is represented by $\operatorname{m} = tan\theta$, where m is the slope of that line which forms $\theta$ angle with the x-axis .