Question

Given, $\sin x=\frac{2}{5}$ and $x\in (0,\frac{\pi }{2})$

Match the following:

• A. $\frac{2}{\sqrt{21}}$
• B. $\frac{\sqrt{21}}{5}$
• C. $\frac{\sqrt{21}}{2}$
• D. $\frac{5}{\sqrt{21}}$

Answer 1

$x$ is an acute angle, so, we will construct an appropriate right triangle and find the ratios.
We will construct the triangle in such a way that $\sin x=\frac{2}{5}$
Using Pythagoras's theorem, we can find the $3^{rd}$ side of the triangle which is equal to $\sqrt{5^2-2^2}=\sqrt{21}$.

We get the triangle,

Now, $\cos x=\frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{\sqrt{21}}{5}$

$\tan x=\frac{\text{Opposite side}}{\text{Hypotenuse}}=\frac{2}{\sqrt{21}}$

$\cot x=\frac{\text{Hypotenuse}}{\text{Opposite side}}=\frac{\sqrt{21}}{2}$

$\sec x=\frac{\text{Hypotenuse}}{\text{Adjacent side}}=\frac{5}{\sqrt{21}}$