Question

If $\tan x=\dfrac{3}{4},0< x< {{90}^{\circ }}$, then what is the value of $\sin x$?
A. $\dfrac{3}{5}$
B. $\dfrac{4}{5}$
C. $\dfrac{12}{25}$
D. $\dfrac{13}{25}$

Answer 1

We explain the function $\arctan \left( m \right)$. We express the inverse function of tan in the form of $\arctan \left( m \right)={{\tan }^{-1}}m$. We find the angle which gives $\tan x=\dfrac{3}{4}$. Thereafter we take the sin ratio of that angle to find the solution. We also use the representation of a right-angle triangle with height and base ratio being $\dfrac{3}{4}$.

Complete step-by-step solution:
This given ratio is $\tan x=\dfrac{3}{4}$. We know $\tan \theta =\dfrac{\text{height}}{\text{base}}$. This gives $x={{\tan }^{-1}}\left( \dfrac{3}{4} \right)$.
We can take the representation of a right-angle triangle with height and base ratio being $\dfrac{3}{4}$ and the angle being $x$. The height and base were considered with respect to that particular angle $x$.

In this case we take $AB=4p$ and keeping the ratio in mind we have $AC=3p$ as the ratio has to be $\dfrac{3}{4}$.
Now we apply the Pythagoras’ theorem to find the length of BC. $B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}$.
So, $B{{C}^{2}}={{\left( 4p \right)}^{2}}+{{\left( 3p \right)}^{2}}=25{{p}^{2}}$ which gives $BC=5p$. Length can’t be negative.
We need to find $\sin \left( {{\tan }^{-1}}\left( \dfrac{3}{4} \right) \right)$ which is equal to $\sin x$.
This ratio gives $\sin x=\dfrac{\text{height}}{\text{hypotenuse}}$. So, $\sin x=\dfrac{AC}{BC}=\dfrac{3p}{5p}=\dfrac{3}{5}$.
Therefore, $\sin x$ is equal to $\dfrac{3}{5}$. The correct option is A.

Note: We can also apply the trigonometric image form to get the value of $\sin x$. It’s given that $\tan x=\dfrac{3}{4}$ and we try to find $\sin x$. We know $\cot x=\dfrac{1}{\tan x}=\dfrac{4}{3}$. We also have $\csc \theta =\sqrt{1+{{\cot }^{2}}\theta }=\sqrt{1+\left( \dfrac{4}{3} \right)}=\dfrac{5}{3}$. Putting the values, we get $\sin x=\dfrac{1}{\csc x}=\dfrac{3}{5}$.