Need Help With Quadrilateral Calculations: LN and KM Angles

AI Thread Summary
The discussion focuses on calculating the angles that lines LN and KM make with the x-axis in a quadrilateral defined by the points K(-2, 7), L(8, 7), M(14, -1), and N(4, -1). The user successfully calculated the angle for line LN as 63.4 degrees using the slope and inverse tangent method. However, they initially struggled with line KM, which has a negative slope, leading to confusion about the angle's representation. It was clarified that when the angle is negative, it can be adjusted by subtracting from 180 degrees to find the correct angle with the x-axis. The discussion emphasizes the importance of understanding the context of angle measurements based on the problem's requirements.
sunny304
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Hey guys, not touched any mathematics in almost 5 years and I am just getting back into it, been going well so far but hit a hurdle with this, I presume its some gaps in my knowledge.

So been at this for about an hour, got the first one, no clue on the second, can anyone offer assistance?K(-2, 7), L(8, 7), M(14, -1), N(4, -1) are the vertices of a quadrilateral.

Calculate the angle that the lines LN and KM make with the x-axis.

For LN I worked out the gradient/slope of the line and used the inverse of tan to come to the solution 63.4 which seems to be correct. However with KM employing the same method doesn't produce the correct result.

Thanks for any help.
 
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sunny304 said:
Hey guys, not touched any mathematics in almost 5 years and I am just getting back into it, been going well so far but hit a hurdle with this, I presume its some gaps in my knowledge.

So been at this for about an hour, got the first one, no clue on the second, can anyone offer assistance?


K(-2, 7), L(8, 7), M(14, -1), N(4, -1) are the vertices of a quadrilateral.

Calculate the angle that the lines LN and KM make with the x-axis.

For LN I worked out the gradient/slope of the line and used the inverse of tan to come to the solution 63.4 which seems to be correct. However with KM employing the same method doesn't produce the correct result.

Thanks for any help.

For KM, the slope is just negative, right? That will give you a negative theta when you use the tan^-1 function. What value did you get?
 
Ah its just clicked, the x-axis can't be more than 180 degrees, so when I get the negative theta, I just subtract from 180 to get the angle. So Simple.

Thank you
 
sunny304 said:
Ah its just clicked, the x-axis can't be more than 180 degrees, so when I get the negative theta, I just subtract from 180 to get the angle. So Simple.

Thank you

Yes, but be careful to look at what the problem question is -- it can vary depending on the question. Sometimes they will want the negative angle as the answer. Other times they will want you to subtract the angle from 180. Just depends on how the question is worded.
 
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