Answer
$\sin A = \frac{2}{\sqrt 5}$
$\cos A = \frac{1}{\sqrt 5}$
$\tan A = 2 $
$\csc A = \frac{\sqrt 5}{2}$
$\sec A $ = $ \sqrt 5 $
$\cot A = \frac{1}{2}$
Work Step by Step
Steps to Answer-
We will use given data about triangle ABC & Pythagoras Theorem to solve for 'c'-
We know that -
$c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem)
$c^{2} = (2)^{2} + (1)^{2}$
$c^{2} = 4 +1$
$c^{2} = 5$
therefore $ c = \sqrt 5$
Now we can write the required T-functions of A using $a=2$ , b = 1 and $ c = \sqrt 5$
$\sin A = \frac{a}{c} = \frac{2}{\sqrt 5}$
$\cos A = \frac{b}{c} = \frac{1}{\sqrt 5}$
$\tan A = \frac{a}{b} =\frac{2}{1}$ = 2
$\csc A = \frac{c}{a} = \frac{\sqrt 5}{2}$
$\sec A = \frac{c}{b} = \frac{\sqrt 5}{1}$ = $ \sqrt 5 $
$\cot A = \frac{b}{a} = \frac{1}{2}$
