the point (5, -2) is on the terminal ray of angle θ, which is in standard position. without evaluating…

Answer:

First, identify $x = 5$, $y=-2$. Then find $r=\sqrt{x^{2}+y^{2}}=\sqrt{5^{2}+(-2)^{2}}=\sqrt{25 + 4}=\sqrt{29}$. The six - trigonometric functions are defined as follows: $\sin\theta=\frac{y}{r}$, $\cos\theta=\frac{x}{r}$, $\tan\theta=\frac{y}{x}$, $\csc\theta=\frac{r}{y}$, $\sec\theta=\frac{r}{x}$, $\cot\theta=\frac{x}{y}$.

Explanation:

Step1: Identify x and y values

Given point $(5,-2)$, so $x = 5$, $y=-2$.

Step2: Calculate r value

Use formula $r=\sqrt{x^{2}+y^{2}}$, substitute $x = 5$ and $y=-2$ to get $r=\sqrt{25 + 4}=\sqrt{29}$.

Step3: Recall trig - function definitions

$\sin\theta=\frac{y}{r}$, $\cos\theta=\frac{x}{r}$, $\tan\theta=\frac{y}{x}$, $\csc\theta=\frac{r}{y}$, $\sec\theta=\frac{r}{x}$, $\cot\theta=\frac{x}{y}$.