Question

which of the following shows a \\(\frac{16\pi}{3}\\) radian angle in standard position? four coordinate plane diagrams with yellow arrows representing angles

Explanation:

Step1: Reduce the angle

To find the coterminal angle of $\frac{16\pi}{3}$, we subtract multiples of $2\pi$ (since $2\pi$ is a full rotation).
First, note that $2\pi=\frac{6\pi}{3}$, and we can find how many full rotations are in $\frac{16\pi}{3}$.
Divide $\frac{16\pi}{3}$ by $2\pi$: $\frac{\frac{16\pi}{3}}{2\pi}=\frac{16\pi}{3}\times\frac{1}{2\pi}=\frac{8}{3}=2+\frac{2}{3}$. So we subtract $2\times2\pi = 4\pi=\frac{12\pi}{3}$ from $\frac{16\pi}{3}$.
$\frac{16\pi}{3}-\frac{12\pi}{3}=\frac{4\pi}{3}$. Wait, no, wait: Wait, $2$ full rotations is $4\pi$, but let's do it step by step. Wait, actually, the formula for coterminal angles is $\theta - 2k\pi$ where $k$ is an integer such that the result is between $0$ and $2\pi$.
Wait, $\frac{16\pi}{3}-2\pi=\frac{16\pi}{3}-\frac{6\pi}{3}=\frac{10\pi}{3}$ (still more than $2\pi$).
$\frac{10\pi}{3}-2\pi=\frac{10\pi}{3}-\frac{6\pi}{3}=\frac{4\pi}{3}$ (still more than $2\pi$? No, $2\pi=\frac{6\pi}{3}$, so $\frac{4\pi}{3}$ is less than $2\pi$? Wait, no, $\frac{4\pi}{3}\approx4.188$, $2\pi\approx6.283$, so $\frac{4\pi}{3}$ is less than $2\pi$. Wait, no, wait, I made a mistake. Wait, $\frac{16\pi}{3}=5\pi+\frac{\pi}{3}$? No, wait, $3\times5 = 15$, so $\frac{16\pi}{3}=5\pi+\frac{\pi}{3}$? No, that's wrong. Wait, $2\pi$ is a full circle. Let's do it correctly:
We know that $2\pi$ radians is 360 degrees, a full circle. So to find the coterminal angle, we can write $\frac{16\pi}{3}=2\pi\times2+\frac{4\pi}{3}$? Wait, no, $2\times2\pi = 4\pi=\frac{12\pi}{3}$, so $\frac{16\pi}{3}-4\pi=\frac{16\pi}{3}-\frac{12\pi}{3}=\frac{4\pi}{3}$. Wait, but $\frac{4\pi}{3}$ is in the third quadrant (between $\pi$ and $\frac{3\pi}{2}$? No, $\pi=\frac{3\pi}{3}$, $\frac{4\pi}{3}$ is between $\pi$ and $\frac{3\pi}{2}$? Wait, no, $\frac{3\pi}{2}=\frac{9\pi}{6}$, $\frac{4\pi}{3}=\frac{8\pi}{6}$, so $\frac{4\pi}{3}$ is between $\pi$ ($\frac{3\pi}{3}$) and $\frac{3\pi}{2}$ ($\frac{9\pi}{6}$)? Wait, maybe I messed up. Wait, let's convert $\frac{16\pi}{3}$ to degrees to check. $\pi$ radians is 180 degrees, so $\frac{16\pi}{3}\times\frac{180}{\pi}=16\times60 = 960$ degrees. Now, subtract 360 degrees until we get an angle between 0 and 360. 960 - 2*360 = 960 - 720 = 240 degrees. 240 degrees is in the third quadrant (180 < 240 < 270). Wait, 240 degrees is $\frac{4\pi}{3}$ radians? Wait, no: 180 degrees is $\pi$, 270 is $\frac{3\pi}{2}$, 240 degrees is $\frac{4\pi}{3}$? Wait, 240/180 = 4/3, so yes, 240 degrees is $\frac{4\pi}{3}$ radians. Wait, but $\frac{4\pi}{3}$ is in the third quadrant (between $\pi$ and $\frac{3\pi}{2}$). Wait, but let's check the coterminal angle again. Wait, maybe I made a mistake in subtracting. Let's do it properly:

The general formula for coterminal angles is $\theta + 2k\pi$, where $k$ is an integer. We want to find $k$ such that $\theta + 2k\pi$ is between $0$ and $2\pi$.

For $\theta=\frac{16\pi}{3}$, let's solve for $k$:

We want $0\leq\frac{16\pi}{3}+2k\pi<2\pi$

Subtract $\frac{16\pi}{3}$: $-\frac{16\pi}{3}\leq2k\pi<2\pi-\frac{16\pi}{3}$

Simplify right side: $2\pi-\frac{16\pi}{3}=\frac{6\pi - 16\pi}{3}=-\frac{10\pi}{3}$

Divide by $2\pi$: $-\frac{8}{3}\leq k<-\frac{5}{3}$

So $k = -3$ (since $k$ must be integer). Then $\theta + 2(-3)\pi=\frac{16\pi}{3}-6\pi=\frac{16\pi}{3}-\frac{18\pi}{3}=-\frac{2\pi}{3}$. Wait, that's negative. Alternatively, maybe I should add $2k\pi$ to get it positive. Wait, maybe my initial approach was wrong. Let's think differently: the angle $\frac{16\pi}{3}$: let's divide 16 by 6 (since $2\pi = 6$ thirds of $\pi$? Wait, no, $2\pi$ is $\frac{6\pi}{3}$. So $\frac{16\pi}{3}=2\times\frac{6\pi}{3}+\frac{4\pi}{3}=2\times2\pi+\frac{4\pi}{3}$. So the coterminal angle is $\frac{4\pi}{3}$, which is in the third quadrant (between $\pi$ and $\frac{3\pi}{2}$). Wait, but $\frac{4\pi}{3}$ is 240 degrees, which is in the third quadrant (x negative, y negative). Wait, but looking at the diagrams:

First diagram: angle in second quadrant (x positive, y positive? No, the yellow arrow is in second quadrant (between 90 and 180 degrees).

Second diagram: angle in fourth quadrant (x positive, y negative).

Third diagram: angle in first quadrant (x positive, y positive).

Fourth diagram: angle in third quadrant (x positive? No, the yellow arrow is below the x-axis, left of y-axis? Wait, the fourth diagram: the yellow arrow is in the third quadrant (x negative? No, the initial side is positive x-axis, and the terminal side is in the third quadrant (below x-axis, left of y-axis? Wait, no, the fourth diagram: the yellow arrow is in the third quadrant (between 180 and 270 degrees), which is where $\frac{4\pi}{3}$ (240 degrees) is. Wait, but let's re-examine the coterminal angle. Wait, maybe I made a mistake. Let's calculate $\frac{16\pi}{3}-2\times2\pi=\frac{16\pi}{3}-4\pi=\frac{16\pi - 12\pi}{3}=\frac{4\pi}{3}$. $\frac{4\pi}{3}$ is 240 degrees, which is in the third quadrant (180 < 240 < 270), so the terminal side should be in the third quadrant. Looking at the diagrams:

First diagram: second quadrant (90 - 180 degrees)

Second diagram: fourth quadrant (270 - 360 degrees)

Third diagram: first quadrant (0 - 90 degrees)

Fourth diagram: third quadrant (180 - 270 degrees)

So the fourth diagram (the last one) shows the angle in the third quadrant, which matches $\frac{4\pi}{3}$ (the coterminal angle of $\frac{16\pi}{3}$). Wait, but let's confirm:

Wait, maybe I messed up the coterminal angle. Let's do it again:

$\frac{16\pi}{3}-3\times2\pi=\frac{16\pi}{3}-6\pi=\frac{16\pi - 18\pi}{3}=-\frac{2\pi}{3}$. The coterminal angle of $-\frac{2\pi}{3}$ is $2\pi-\frac{2\pi}{3}=\frac{4\pi}{3}$, which is the same as before. So $\frac{4\pi}{3}$ is in the third quadrant. So the diagram with the terminal side in the third quadrant (the fourth one) is the correct one.

Wait, but let's check the diagrams again. The fourth diagram: the yellow arrow is in the third quadrant (below the x-axis, to the left of the y-axis? Wait, no, the initial side is positive x-axis, and the terminal side is in the third quadrant (between 180 and 270 degrees), which is where $\frac{4\pi}{3}$ is. So that's the correct one.

Step2: Identify the quadrant

$\frac{4\pi}{3}$ radians is in the third quadrant (since $\pi<\frac{4\pi}{3}<\frac{3\pi}{2}$), so the terminal side should be in the third quadrant. Looking at the options, the fourth diagram (the last one) has the terminal side in the third quadrant.

Answer:

The diagram with the terminal side in the third quadrant (the last one among the four given diagrams) shows the $\frac{16\pi}{3}$ radian angle in standard position. (Assuming the fourth diagram is the one with the terminal side in the third quadrant, as per the description: "the yellow arrow is in the third quadrant" – based on the last diagram's description where the yellow arrow is below the x-axis and left of the y-axis? Wait, the last diagram: "the yellow arrow is" – looking at the user's diagram, the last one has the yellow arrow in the third quadrant (between 180 and 270 degrees), so that's the correct one.