# Homework Help: Standard position angle.

1. Mar 27, 2009

### sandynair

I have a question can someone help me?

If I want to determine the relates acute angle associated with each of the following standard pposition angles : 12/7 pi what is the answer and can someone show me as a drawing?

thanks

2. Mar 28, 2009

### HallsofIvy

The "standard position" angle is between 0 and $\pi/2$ radians. 12/7= 1 and 5/7. Since 5/7 is larger than 1/2, the standard angle is $\pi- (5/7)pi= (1- 5/7)\pi= (2/7)\pi$.

The angle itself, drawn on an xy-coordinate system, is in the fourth quadrant, measured from the negative y-axis. The standard position is that same angle but in the first quadrant, measured from positive x-axis.

3. Mar 28, 2009

### sandynair

how is 12/7 = 1

4. Mar 28, 2009

### Staff: Mentor

It's not. He wrote 12/7 = 1 and 5/7, meaning 1 5/7.

5. Mar 30, 2009

### meiso

Just to clarify, an angle in standard position is simply one whose vertex lies at the origin and whose initial side coincides with the positive x-axis. The terminal side of the angle can rotate clockwise or counterclockwise and lie in any quadrant.

The original poster was saying he was required to find the related acute angle (reference angle) associated with the following standard position angles (and he only listed 12pi/7 as one of these standard position angles).

The easiest way to find the reference angle for 12pi/7 is, once you know it terminates in the 4th quadrant (by noticing 12/7 is greater than 3/2), simply subtract 2pi - 12pi/7 and you will get 2pi/7.

As a general note, subtracting the measure of your quadrant four angle from 2pi is the way to get the reference angle for any standard position angle terminating in quadrant four.

For those that terminate in:
QI: ref angle = std. pos. angle
QII: ref angle = pi - (std. pos. angle)
QIII: ref angle = (std. pos. angle) - pi
QIV: ref angle = 2pi - (std. pos. angle)

Last edited: Mar 30, 2009
6. Mar 31, 2009

^

Nice post.