## Thinking Mathematically (6th Edition)

$\sin{A} = \frac{3}{5}$ $\cos{A} = \frac{4}{5}$ $\tan{A} =\frac{3}{4}$
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: $\sin{A} = \frac{9}{15} = \frac{3}{5}$ $\cos{A} = \frac{12}{15} = \frac{4}{5}$ $\tan{A} = \dfrac{9}{12}=\frac{3}{4}$