#### Answer

$\sin{A} = \frac{3}{5}$
$\cos{A} = \frac{4}{5}$
$\tan{A} =\frac{3}{4}$

#### Work Step by Step

RECALL:
In a right triangle,
(1) $c^2=a^2+b^2$ where c = hypotenuse, a and b are the legs.
(2)
$\sin{A} = \dfrac{\text{length of opposite side}}{\text{length of hypotenuse}}$
$\cos{A} = \dfrac{\text{length of adjacent side}}{\text{length of hypotenuse}}$
$\tan{A} = \dfrac{\text{length of opposite side}}{\text{length of adjacent side}}$
Solve for the length of the hypotenuse using formula (1) to obtain:
$c^2= a^2 + b^2
\\c^2 = 12^2 + 9^2
\\c^2 = 144 + 81
\\c^2 = 225
\\c^2 = 15^2
\\c = \sqrt{15^2}
\\c = 15$
Use the formulas in (2) above to obtain:
$\sin{A} = \frac{9}{15} = \frac{3}{5}$
$\cos{A} = \frac{12}{15} = \frac{4}{5}$
$\tan{A} = \dfrac{9}{12}=\frac{3}{4}$