Angle on coordinate axis, related acute angle

In summary, the conversation discusses finding two possible locations for the terminal arm of angle A, as well as two possible values for the measure of angle A and its related acute angle. The correct triangles are drawn in quadrants II and IV, with x and y values being switched depending on the quadrant to satisfy the given value of tan A.
  • #1
pbonnie
92
0

Homework Statement


Tan A = -2/6
a) Draw two possible locations on the coordinate axis for the terminal arm of angle A
b) Find two possible values for the measure of angle A and the related acute angle.


Homework Equations


c^2 = a^2 + b^2
SOHCAHTOA


The Attempt at a Solution


I know that the two possible locations are in quadrants II and IV. But I'm not sure how they're supposed to be drawn? I attached an image of how I think it's supposed to be, but I'm not sure if I did it right.
Also, I'm not sure what angle A represents? Is it the angle IN the triangle? or the angle outside of it?
 
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  • #2
Picture
 

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  • #3
pbonnie said:

Homework Statement


Tan A = -2/6
a) Draw two possible locations on the coordinate axis for the terminal arm of angle A
b) Find two possible values for the measure of angle A and the related acute angle.


Homework Equations


c^2 = a^2 + b^2
SOHCAHTOA


The Attempt at a Solution


I know that the two possible locations are in quadrants II and IV. But I'm not sure how they're supposed to be drawn? I attached an image of how I think it's supposed to be, but I'm not sure if I did it right.
Unfortunately, the triangles are not drawn correctly. The way you have it, tan A = -6/2.

pbonnie said:
Also, I'm not sure what angle A represents? Is it the angle IN the triangle? or the angle outside of it?
For angles in standard position, the initial side lies on the positive x-axis. You measure them by going counterclockwise until you reach the terminal side. So in quadrant I, the angle is in the triangle, but for the other quadrants, the angle is NOT in the triangle.
 
  • #4
Okay thanks, so would the triangles look more like this? (image attached)

And does that mean the related acute angle is the angle IN the triangle, and the angle A (in the 2nd quadrant) would be 180 - (related acute angle)? and the angle A (in the 4th quadrant) would be 360 - (related acute angle)?
 

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  • #5
pbonnie said:
Okay thanks, so would the triangles look more like this? (image attached)
Still incorrect, I'm afraid. Take the first diagram you uploaded and just switch the 6's and 2's (but leave the negatives alone). Remember that tan θ = y/x.

pbonnie said:
And does that mean the related acute angle is the angle IN the triangle, and the angle A (in the 2nd quadrant) would be 180 - (related acute angle)? and the angle A (in the 4th quadrant) would be 360 - (related acute angle)?

Yes.
 
  • #6
The triangle in Quadrant IV is flipped.
 
  • #7
Oh okay, I think I understand. Like this?

I'm just not sure why that happens? Why can x be -6 or 6 and y be -2 or 2?
 

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  • #8
The b in the 4th quadrant is in the wrong spot, meant to put it along x-axis
 
  • #9
6* not b
 
  • #10
pbonnie said:
Oh okay, I think I understand. Like this?

I'm just not sure why that happens? Why can x be -6 or 6 and y be -2 or 2?
Yes (after taking into account the misplacement of 6 in Q IV), but I don't understand your question. The way your latest diagram is drawn , in Quadrant II x = -6 and y = 2, and in Quadrant IV, x = 6 and y = -2. :confused:
 
  • #11
eumyang said:
Yes (after taking into account the misplacement of 6 in Q IV), but I don't understand your question. The way your latest diagram is drawn , in Quadrant II x = -6 and y = 2, and in Quadrant IV, x = 6 and y = -2. :confused:

That is the correct triagnles though? That's what I mean, I just don't understand why the values are switched? (from x being negative to positive, and same with y)
 
  • #12
pbonnie said:
That is the correct triagnles though? That's what I mean, I just don't understand why the values are switched? (from x being negative to positive, and same with y)
You have [itex]\tan A = -\frac{2}{6}[/itex], and you know that [itex]\tan \theta = \frac{y}{x}[/itex]. If you have a fraction that is negative, then either the numerator or the denominator is negative, but not both. So these are equivalent:
[itex]-\frac{2}{6} = \frac{-2}{6} = \frac{2}{-6}[/itex]
Therefore, we draw triangles in Quadrants II and IV. In Quadrant II, x is negative and y is positive. In Quadrant IV, x is positive and y is negative.

Does that answer your question? If not, then I still don't know what you are asking.
 
  • #13
Oh okay. Yes, that explains it. Thank you!
 

1. What is an angle on a coordinate axis?

An angle on a coordinate axis is a measurement of the rotation between two intersecting lines, one horizontal and one vertical, on a coordinate plane. It is typically measured in degrees or radians.

2. How do you find the acute angle related to an angle on a coordinate axis?

The acute angle related to an angle on a coordinate axis can be found by subtracting the given angle from 90 degrees. This is because an acute angle is always less than 90 degrees.

3. What is the purpose of finding the related acute angle?

The purpose of finding the related acute angle is to simplify the measurement and make it easier to work with. In many cases, the related acute angle will be easier to visualize and manipulate than the given angle on the coordinate axis.

4. Can the related acute angle be negative?

No, the related acute angle cannot be negative. Acute angles are always positive, meaning they are measured in a counterclockwise direction from the positive x-axis.

5. How is an angle on a coordinate axis used in real-life applications?

An angle on a coordinate axis is used in various fields, such as engineering, surveying, and physics, to measure and analyze the position and orientation of objects in space. It is also used in navigation, map-making, and computer graphics.

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