## Trigonometry 7th Edition

$\sin A = \frac{2}{3}$ $\cos A = \frac{\sqrt 5}{3}$ $\tan A =\frac{2}{\sqrt 5}$ $\csc A = \frac{3}{2}$ $\sec A = \frac{3}{\sqrt 5}$ $\cot A = \frac{\sqrt 5}{2}$
Steps to Answer- We will use given data about triangle ABC &amp; 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 $c = 3$ $\sin A = \frac{a}{c} = \frac{2}{3}$ $\cos A = \frac{b}{c} = \frac{\sqrt 5}{3}$ $\tan A = \frac{a}{b} =\frac{2}{\sqrt 5}$ $\csc A = \frac{c}{a} = \frac{3}{2}$ $\sec A = \frac{c}{b} = \frac{3}{\sqrt 5}$ $\cot A = \frac{b}{a} = \frac{\sqrt 5}{2}$