Question 1089376: Find cos(z) and sin(z) if z is an angle in quadrant III (in standard position) and the terminal side of angle z is parallel to the line 3x+4y=12.
I just want to make sure I'm thinking about this correctly:
The definition of an angle in standard position is that the vertex is at (0,0) and the the x-axis is one end of the angle. Now, I'm thinking that the line with which the terminal side coincides MUST contain the point (0,0) because the terminal side of an angle must be connected to it's vertex and by definition the vertex of this angle is at (0,0). My problem lies with the fact that if the angle is in quadrant III (which if I remember correctly is the bottom left corner of the Cartesian plane) then the terminal side must also be in quadrant III and the line that contains that terminal side must be parallel to 3x+4y=12 and must contain the point (0,0). But this is impossible, because if a line has the point (0,0) that is parallel to 3x+4y=12, it will have the slope -(3/4), from which we get that the line that contains the terminal side is just y=-(3/4)x, but this line doesn't contain any points in quadrant III (because when x is negative y is positive and thus in quadrant II). Therefore, angle z CANNOT have a terminal side in quadrant III parallel to the line 3x+4y=12 because there exists no parallel line to 3x+4y=12 that contains the point (0,0) that also contains points in quadrant III.
So I'm getting that this question is impossible. Is my reasoning flawed here? Am I missing something?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
You are correct. The problem as stated is impossible to answer. There must be a typo somewhere.
-------------------------------------------------------------------------------------
Refer to the figure below
(Image generated by GeoGebra which is free graphing software)
The graph of 3x+4y = 12 is shown in red. It passes through (0,3) which is the y intercept and (4,0) which is the x intercept
The angle is composed of two segments (green and purple). The terminal side is the purple segment.
Imagine that B and C are fixed and cannot be moved, but A is allowed to roam around as long as both x and y coordinates are negative together.
In this example, the angle is 225 degrees.
Based on the figure, we see that line AB has a positive slope while the slope of 3x+4y = 12 is negative.
No matter where A is placed (according to the conditions above), we cannot make AB have a negative slope.
So it's impossible for AB to be parallel to 3x+4y = 12 (as the slopes must be equal for parallel lines to occur).
Which means that it's impossible for the terminal side to be parallel to 3x+4y = 12.
Much of what I'm saying, if not all of it, is simply a paraphrase of what you already wrote. So you have the correct logic and reasoning. I'm simply offering a slightly different viewpoint with an example drawing. Let me know if this helps or not. Thank you.
|
|
|
