Answer
$\sin A = \frac{12}{13}$
$\\cos A = \frac{5}{13}$
$\\tan A = \frac{12}{5}$
$\\csc A = \frac{13}{12}$
$\\\sec A = \frac{13}{5}$
$\\\cot A = \frac{5}{12}$
Work Step by Step
We will use given data about triangle ABC & Pythagoras Theorem to solve for 'a'-
We know that -
$c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem)
Therefore -
$a^{2} =c^{2} - b^{2}$
$a^{2} = (13)^{2} - (5)^{2}$
$a^{2} = 169 - 25$
$a^{2} = 144$
therefore $ a = \sqrt (144)$
a = 12
Now we can write the required T-functions of A and B using $a=12$ , b = 5 and c = 13
$\sin A = \frac{a}{c} = \frac{12}{13}$
$\\cos A = \frac{b}{c} = \frac{5}{13}$
$\\tan A = \frac{a}{b} = \frac{12}{5}$
$\\csc A = \frac{c}{a} = \frac{13}{12}$
$\\\sec A = \frac{c}{b} = \frac{13}{5}$
$\\\cot A = \frac{b}{a} = \frac{5}{12}$
