Question

sketch the least positive angle θ and find the values of the six trigonometric functions of θ if the terminal side of an angle θ in standard position is defined by -√10x + y = 0, x ≤ 0. which of the graphs shows -√10x + y = 0, where x ≤ 0? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. sin θ = (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.) b. the function is undefined. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. cos θ = (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Rewrite the line equation

Rewrite $-\sqrt{10}x + y=0$ as $y = \sqrt{10}x$. Since $x\leq0$, the terminal - side of the angle $\theta$ is in the second - quadrant. Let $x=-1$, then $y =-\sqrt{10}$.

Step2: Calculate the radius $r$

Use the formula $r=\sqrt{x^{2}+y^{2}}$. Substitute $x = - 1$ and $y=-\sqrt{10}$ into it, we get $r=\sqrt{(-1)^{2}+(-\sqrt{10})^{2}}=\sqrt{1 + 10}=\sqrt{11}$.

Step3: Calculate $\sin\theta$

By the definition $\sin\theta=\frac{y}{r}$, substituting $y =-\sqrt{10}$ and $r=\sqrt{11}$, we have $\sin\theta=\frac{-\sqrt{10}}{\sqrt{11}}=-\frac{\sqrt{110}}{11}$.

Step4: Calculate $\cos\theta$

By the definition $\cos\theta=\frac{x}{r}$, substituting $x=-1$ and $r = \sqrt{11}$, we get $\cos\theta=-\frac{1}{\sqrt{11}}=-\frac{\sqrt{11}}{11}$.

Step5: Calculate $\tan\theta$

By the definition $\tan\theta=\frac{y}{x}$, substituting $x=-1$ and $y =-\sqrt{10}$, we have $\tan\theta=\sqrt{10}$.

Step6: Calculate $\csc\theta$

Since $\csc\theta=\frac{r}{y}$, substituting $y =-\sqrt{10}$ and $r=\sqrt{11}$, we get $\csc\theta=-\frac{\sqrt{11}}{\sqrt{10}}=-\frac{\sqrt{110}}{10}$.

Step7: Calculate $\sec\theta$

Since $\sec\theta=\frac{r}{x}$, substituting $x=-1$ and $r=\sqrt{11}$, we have $\sec\theta=-\sqrt{11}$.

Step8: Calculate $\cot\theta$

Since $\cot\theta=\frac{x}{y}$, substituting $x=-1$ and $y =-\sqrt{10}$, we get $\cot\theta=\frac{-1}{-\sqrt{10}}=\frac{\sqrt{10}}{10}$.

For the graph of $y=\sqrt{10}x$ with $x\leq0$, the line has a positive slope and is in the second - quadrant.

Answer:

For the graph question: The line $y = \sqrt{10}x$ with $x\leq0$ has a positive slope and lies in the second - quadrant. Without seeing the actual graphs, we know it should be a line passing through the origin with positive slope in the second - quadrant part.
For $\sin\theta$: A. $\sin\theta=-\frac{\sqrt{110}}{11}$
For $\cos\theta$: A. $\cos\theta=-\frac{\sqrt{11}}{11}$
For $\tan\theta$: A. $\tan\theta=\sqrt{10}$
For $\csc\theta$: A. $\csc\theta=-\frac{\sqrt{110}}{10}$
For $\sec\theta$: A. $\sec\theta=-\sqrt{11}$
For $\cot\theta$: A. $\cot\theta=\frac{\sqrt{10}}{10}$