Question

If $\tan x=3.5$, how do you find the value of $x$ to the nearest tenth ?

Answer 1

Hint:We explain the function $arc\tan \left( x \right)$. We express the inverse function of tan in the form of $arc\tan \left( x \right)={{\tan }^{-1}}x$. We draw the graph of $arc\tan \left( x \right)$ and the line $x=3.5$ to find the intersection point as the solution of $\tan x=3.5$.

Complete step by step answer:
We have $\tan x=3.5=\dfrac{7}{2}$ which gives $x={{\tan }^{-1}}\left( \dfrac{7}{2} \right)$.The given expression is the inverse function of trigonometric ratio tan.The arcus function represents the angle which on ratio tan gives the value.So, $arc\tan \left( x \right)={{\tan }^{-1}}x$. If $arc\tan \left( x \right)=\alpha $ then we can say $\tan \alpha =x$.

Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of $2\pi $.The general solution for that value where $\tan \alpha =x$ will be $n\pi +\alpha ,n\in \mathbb{Z}$.But for $arc\tan \left( x \right)$, we won’t find the general solution. We use the principal value. For ratio tan we have $-\dfrac{\pi }{2}\le arc\tan \left( x \right)\le \dfrac{\pi }{2}$.The graph of the function is,

$arc\tan \left( x \right)=\alpha $ gives the angle $\alpha $ behind the ratio.We now place the value of $x=\dfrac{7}{2}$ in the function of $arc\tan \left( x \right)$.Let the angle be $\theta $ for which $arc\tan \left( \dfrac{7}{2} \right)=\theta $. This gives $\tan \theta =\dfrac{7}{2}$. Putting the value in the graph of $arc\tan \left( x \right)$, we get $\theta =74.05$. For this we take the line of $x=\dfrac{7}{2}$ and see the intersection of the line with the graph $arc\tan \left( x \right)$.

Therefore, the value of $arc\tan \left( \dfrac{4}{3} \right)$ is ${{74.05}^{\circ }}$.

Note: First note that the value $\dfrac{7}{2}$ looks suspiciously like it was intended to be an angle but the argument of the $arc\tan \left( x \right)$ function is not an angle. The representation will be the right-angle triangle with base 2 and height 7 and the angle being $\theta $.