Question

which of the following shows a $-\frac{5pi}{6}$ radian angle in standard position? four coordinate plane diagrams with purple angle rays are shown, but their specific details are omitted here as per the instruction to focus on text and formulas

Explanation:

Step1: Recall negative angle direction

Negative angles in standard position are measured clockwise from the positive x - axis. The angle is $-\frac{5\pi}{6}$ radians. Let's convert this to a positive angle by adding $2\pi$: $-\frac{5\pi}{6}+2\pi=\frac{- 5\pi + 12\pi}{6}=\frac{7\pi}{6}$, but we can also analyze the clockwise rotation. The magnitude of the angle is $\frac{5\pi}{6}\approx150^{\circ}$.

Step2: Analyze quadrants

A clockwise rotation of $\frac{5\pi}{6}$ radians (or $150^{\circ}$) from the positive x - axis will land the terminal side in the third quadrant? Wait, no. Wait, a clockwise rotation of $90^{\circ}$ ($\frac{\pi}{2}$ radians) takes us to the negative y - axis, and a clockwise rotation of $180^{\circ}$ ($\pi$ radians) takes us to the negative x - axis. $\frac{5\pi}{6}\approx150^{\circ}$, which is more than $90^{\circ}$ ($\frac{\pi}{2}$) and less than $180^{\circ}$ ($\pi$). So a clockwise rotation of $150^{\circ}$ from the positive x - axis: starting at positive x - axis, rotating clockwise $90^{\circ}$ (to negative y - axis), then rotating an additional $60^{\circ}$ ($\frac{\pi}{3}$ radians) towards the negative x - axis. Wait, no, $\frac{5\pi}{6}=\pi-\frac{\pi}{6}$, so a clockwise rotation of $\frac{5\pi}{6}$ radians: the reference angle (the acute angle with the x - axis) is $\frac{\pi}{6}$ ($30^{\circ}$), and the terminal side is in the third quadrant? Wait, no. Wait, when we rotate clockwise, the first quadrant (clockwise from positive x - axis: 0 to $- 90^{\circ}$), second quadrant (clockwise from positive x - axis: $- 90^{\circ}$ to $- 180^{\circ}$), third quadrant (clockwise from positive x - axis: $- 180^{\circ}$ to $- 270^{\circ}$), fourth quadrant (clockwise from positive x - axis: $- 270^{\circ}$ to $- 360^{\circ}$). Wait, $-\frac{5\pi}{6}\approx - 150^{\circ}$, which is between $- 90^{\circ}$ and $- 180^{\circ}$, so it is in the third quadrant? Wait, no. Wait, the standard position: positive angles counter - clockwise, negative angles clockwise. So:
- $0$ to $\frac{\pi}{2}$ (counter - clockwise): first quadrant.
- $\frac{\pi}{2}$ to $\pi$ (counter - clockwise): second quadrant.
- $\pi$ to $\frac{3\pi}{2}$ (counter - clockwise): third quadrant.
- $\frac{3\pi}{2}$ to $2\pi$ (counter - clockwise): fourth quadrant.

For negative angles (clockwise):
- $0$ to $-\frac{\pi}{2}$ (clockwise): fourth quadrant.
- $-\frac{\pi}{2}$ to $-\pi$ (clockwise): third quadrant.
- $-\pi$ to $-\frac{3\pi}{2}$ (clockwise): second quadrant.
- $-\frac{3\pi}{2}$ to $- 2\pi$ (clockwise): first quadrant.

Since $-\frac{5\pi}{6}$ is between $-\frac{\pi}{2}$ ($- 90^{\circ}$) and $-\pi$ ($- 180^{\circ}$), it is in the third quadrant? Wait, no. Wait, $-\frac{\pi}{2}=-\frac{3\pi}{6}$, $-\pi = -\frac{6\pi}{6}$. So $-\frac{5\pi}{6}$ is between $-\frac{3\pi}{6}$ and $-\frac{6\pi}{6}$, so it is in the third quadrant (clockwise from positive x - axis: between $- 90^{\circ}$ and $- 180^{\circ}$). Now let's look at the graphs:

- The first graph: terminal side in third quadrant (below negative x - axis? Wait, no, the first graph's terminal side is in the third quadrant (between negative x - axis and negative y - axis? Wait, the second graph: terminal side in second quadrant (counter - clockwise) or third? Wait, no. Wait, let's re - express:

Wait, $-\frac{5\pi}{6}$ radians: to find the standard position, start at positive x - axis, rotate clockwise $\frac{5\pi}{6}$ radians. $\frac{5\pi}{6}=150^{\circ}$, so clockwise $150^{\circ}$. A clockwise $90^{\circ}$ is to negative y - axis, then another $60^{\circ}$ towards negative x - axis. So the terminal side is in the third quadrant, making an angle of $30^{\circ}$ with the negative x - axis (since the reference angle is $\frac{\pi}{6}$). Now looking at the options:

- Option 1: Terminal side in third quadrant (between negative x - axis and negative y - axis)
- Option 2: Terminal side in second quadrant (counter - clockwise)
- Option 3: Terminal side in fourth quadrant
- Option 4: Terminal side in first quadrant

Wait, maybe I made a mistake. Wait, $-\frac{5\pi}{6}$ radian: let's convert to degrees: $-\frac{5\pi}{6}\times\frac{180^{\circ}}{\pi}=- 150^{\circ}$. So a $- 150^{\circ}$ angle: starting at positive x - axis, rotating clockwise $150^{\circ}$. So the terminal side is in the third quadrant (because $- 90^{\circ}>- 150^{\circ}>- 180^{\circ}$). The first graph: the terminal side is in the third quadrant (below the negative x - axis? Wait, no, the first graph's terminal side is in the third quadrant (between negative x - axis and negative y - axis). Wait, but let's check the reference angle. The reference angle for $-\frac{5\pi}{6}$ is $\frac{\pi}{6}$ (since $\pi-\frac{5\pi}{6}=\frac{\pi}{6}$ for the positive counterpart, but for negative, the reference angle is also $\frac{\pi}{6}$). So the terminal side makes a $30^{\circ}$ angle with the negative x - axis (since $180^{\circ}-150^{\circ}=30^{\circ}$). So the first graph: if the terminal side is in the third quadrant, with a reference angle of $30^{\circ}$ (i.e., making $30^{\circ}$ with negative x - axis), that's correct. Wait, maybe the first graph is the one. Wait, no, let's think again. Wait, $-\frac{5\pi}{6}$ is equivalent to $2\pi-\frac{5\pi}{6}=\frac{7\pi}{6}$, which is in the third quadrant (counter - clockwise $\frac{7\pi}{6}$ radians). So the terminal side of $-\frac{5\pi}{6}$ (clockwise) and $\frac{7\pi}{6}$ (counter - clockwise) is the same. So the terminal side is in the third quadrant, making an angle of $\frac{\pi}{6}$ with the negative x - axis. So the first graph (the first option) has the terminal side in the third quadrant, which matches the terminal side of $-\frac{5\pi}{6}$ radians.

Wait, maybe I messed up the quadrant. Let's take a positive angle of $\frac{7\pi}{6}$ (which is coterminal with $-\frac{5\pi}{6}$). $\frac{7\pi}{6}$ is in the third quadrant (since $\pi<\frac{7\pi}{6}<\frac{3\pi}{2}$). So the terminal side of $-\frac{5\pi}{6}$ is the same as $\frac{7\pi}{6}$, so it's in the third quadrant. So the first graph (the top - most graph) has the terminal side in the third quadrant, so that's the correct one.

Answer:

The first graph (the one with the terminal side in the third quadrant, between the negative x - axis and negative y - axis) shows a $-\frac{5\pi}{6}$ radian angle in standard position. (Assuming the first option is the one with the terminal side in the third quadrant as per the visual description)