$a=\cos20^o$
$b=\sin20^o$
$c=\tan20^o$
Compare $a,b$ and $c$
I could figure out that $a=\cos20^o=\sin70^o$, then $b \lt a$.
I can't put $\tan20^0$ anywhere in the inequality. Usually, in these type of problems, $\tan{x^o}$ would be greater than $45^o$ , so that I can compare it with the others, knowing that it would be greater than $1$.
How do I compare $a$ and $c$?
$\cos 30^\circ$ and $\tan 30^\circ$ can be memorized without using a calculator (see https://math.stackexchange.com/a/1554126/645472).
$\cos 30^\circ = \frac{\sqrt{3}}{2}$ is larger than $\tan 30^\circ = \frac{1}{\sqrt{3}}$ (you can see this by multiplying both with $\sqrt{3}$). Moreover, $\cos \theta$ is decreasing for $\theta$ going from $0^\circ$ to $90^\circ$, whereas $\tan \theta$ is increasing on that interval, so $$ a = \cos 20^\circ > \cos 30^\circ > \tan 30^\circ > \tan 20^\circ = c $$ Combining this with what you already found ($b<a$) and the answer of @Anadactothe ($c>b$), you get $$ a > c > b $$
Use the appropriate half-angle identity to render $\cos^2 20°=(1+\cos 40°)/2=(1+\sin 50°)/2$. Then since $1>\sin 50°>\sin 20°$, conclude that $\cos^2 20° > \sin 20°$ and therefore $\cos 20° > \sin 20°/\cos 20°$.
$c=\frac{b}{a}>b$, while $b<\sin 30^\circ=\frac{1}{2}\implies \frac{c}{a}=\frac{b}{1-b^2}<\frac{2}{3}$. This uses the fact that $\frac{x}{1-x^2}=\frac{1}{2}(\frac{1}{1-x}-\frac{1}{1+x})$ increases on $[0,\,1]$.
For this question, all you need is to know is the quotient identity of tangent, that is $tan(\theta) =\frac{sin(\theta)}{cos(\theta)}$
We know that $cos(0)$ is 1 and is decreasing on the interval $(0,\pi)$, so $cos(20) < 1$
Considering that $tan(20) =\frac {sin(20)}{cos(20)}$, and $cos(20) < 1$, it follows that $tan(20) > sin(20)$ because dividing by a number less than one increases the absolute value of an expression, and looking at the unit circle, sine, cosine, and tangent are positive in quadrant 1.
Therefore, cos(20) > tan(20) > sin(20)
