Chapter 1 - Section 1.3 - Definition I: Trigonometric Functions - 1.3 Problem Set - Page 33: 55

$\sin\theta=\frac{12}{13}$ $\cos\theta=\frac{5}{13}$ $\tan\theta=\frac{12}{5}$

$\sin\theta=\frac{y}{r}$ $\sin\theta=\frac{12}{13}$ Substitute into $x^2+y^2=r$ $x=\pm5$ So x=5 because $\theta$ terminates in the first quadrant and can only be positive. Knowing x, we can now substitute it into the rest of the trigonometric functions. $\cos\theta=\frac{5}{13}$ $\tan\theta=\frac{12}{5}$