Answer
$\sin A = \frac{2}{3}$
$\cos A = \frac{\sqrt 5}{3}$
$\tan A =\frac{2}{\sqrt 5}$
$\csc A = \frac{3}{2}$
$\sec A = \frac{3}{\sqrt 5}$
$\cot A = \frac{\sqrt 5}{2}$
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)
$c^{2} = (2)^{2} + (\sqrt 5)^{2}$
$c^{2} = 4 +5$
$c^{2} =9$
therefore $ c = \sqrt 9$ = 3
Now we can write the required T-functions of A using $a=2$ , $b =\sqrt 5$ and $ c = 3$
$\sin A = \frac{a}{c} = \frac{2}{3}$
$\cos A = \frac{b}{c} = \frac{\sqrt 5}{3}$
$\tan A = \frac{a}{b} =\frac{2}{\sqrt 5}$
$\csc A = \frac{c}{a} = \frac{3}{2}$
$\sec A = \frac{c}{b} = \frac{3}{\sqrt 5}$
$\cot A = \frac{b}{a} = \frac{\sqrt 5}{2}$
