## Precalculus (6th Edition) Blitzer

a) The ratio that expresses $\sin \theta$ is $\frac{y}{r}$. b) The ratio for figure (b) is $\frac{y}{r}=\frac{4}{5}$, that is, positive.
(a) In this figure, both x and y coordinates are positive as they lie in the first quadrant; the ordinate is y and abscissa is x. $\sin \theta =\frac{y}{r}$ (b) In the second figure, we calculate the value of r as follows: \begin{align} & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ & r=\sqrt{{{\left( -3 \right)}^{2}}+{{4}^{2}}} \\ & r=\sqrt{9+16} \\ & r=\sqrt{25} \\ & r=5 \\ & \frac{y}{r}=\frac{4}{5} \end{align} Thus, the above ratio obtained is positive.