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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 \[3\;,Find\;\theta .\]

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Last updated date: 26th Jul 2024
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Answer
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Hint: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 .